Probability

P(A $ \cup $ B) = P(A) + P(B) – P(A $ \cup $ B)

$ \Rightarrow $ 0.8 = 0.6 + 0.4 – P(A $ \cap $ B)

$ \Rightarrow $ P(A $ \cap $ B) = 0.2

P(A$ \cup $B$ \cup $C) = P(A) + P(B) + P(C) – P(A $ \cap $ B) – P(B $ \cap $ C) –P(C $ \cap $ A) + P(A $ \cap $ B $ \cap $ C)

$ \Rightarrow $ $\alpha $ = 0.6 + 0.4 + 0.5 - 0.2 - $\beta $ - 0.3 + 0.2

$ \Rightarrow $ $\alpha $ + $\beta $ = 1.2

$ \Rightarrow $ $\alpha $ = 1.2 - $\beta $

Given, 0.85 $ \le \alpha \le $ 0.95

$ \Rightarrow $ 0.85 $ \le $ 1.2 - $\beta $ $ \le $ 0.95

$ \Rightarrow $ 0.25 $ \le \beta \le $ 0.35

Out of 11 consecutive natural numbers either 6 even and 5 odd numbers or 5 even and 6 odd numbers.

Let, E = Even
O = Odd

Case-1 :

E, O, E, O, E, O, E, O, E, O, E

2b = a + c $ \Rightarrow $ Even

$ \Rightarrow $ Both a and c should be either even or odd.

P = ${{{}^6{C_2} + {}^5{C_2}} \over {{}^{11}{C_3}}}$ = ${5 \over {33}}$

Case -2 :

O, E, O, E, O, E, O, E, O, E, O

P = ${{{}^5{C_2} + {}^6{C_2}} \over {{}^{11}{C_3}}}$ = ${5 \over {33}}$

Total probability = ${1 \over 2} \times {5 \over {33}}$ + ${1 \over 2} \times {5 \over {33}}$ = ${5 \over {33}}$

Sum total 6 = {(1,5)(2,4)(3,3)(4,2)(5,1)}

$P(6) = {5 \over 36}$

Sum total 7 = {(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)}

$\,P(7) = {6 \over {36}}$ = ${1 \over 6}$

Game ends and A wins if A throws 6 in 1^st throw or A don’t throw 6 in 1^st throw, B don’t throw 7 in 1^st throw and then A throw 6 in his 2 ^nd chance and so on.

P(A) = A + $\overline A \overline B A$ + $\overline A \overline B \overline A \overline B A$

= ${5 \over {36}} + \left( {{{31} \over {36}}} \right)\left( {{{30} \over {36}}} \right)\left( {{5 \over {36}}} \right) + $ ..... $\infty $

= ${{{5 \over {36}}} \over {1 - \left( {{{31} \over {36}}} \right)\left( {{{30} \over {36}}} \right)}}$

= ${{5 \times 36} \over {36 \times 36 - 31 \times 30}}$

= ${30 \over {61}}$

Sample space = 9 $ \times $ 10⁴

Case - I

Out of exactly two digits selected one is zero then favourable cases = ${}^9{C_1}({2^4} - 1)$

Case - II

Both selected digits are non-zero then favourable cases = ${}^9{C_2}({2^5} - 2)$

Probability = ${{9({2^4} - 1) + {{9.8} \over 2}({2^5} - 2)} \over {9 \times {{10}^4}}}$

$ = {{15 + 120} \over {{{10}^4}}} = {{135} \over {{{10}^4}}}$

Let A is the event for getting score a multiple of 4.

So, A = { (1, 3), (3, 1), (2, 2), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4), (6, 6) } = 09

n(A) = 9

B : Score of 4 has appeared at least once.

B = {(4, 4)}

So, Required probability = ${1 \over 9}$

Given E₁ , E₂ , E₃ are pairwise indepedent events

so P(E₁ $ \cap $ E₂ ) = P(E₁ ).P(E₂ )

and P(E₂ $ \cap $ E₃ ) = P(E₂ ).P(E₃ )

and P(E₃ $ \cap $ E₁ ) = P(E₃ ).P(E₁ )

and P(E₁ $ \cap $ E₂ $ \cap $ E₃) = 0

Now $P\left( {{{E_2^C \cap E_3^C} \over {{E_1}}}} \right)$

$ = {{P\left[ {{E_1} \cap \left( {E_2^C \cap E_3^C} \right)} \right]} \over {P\left( {{E_1}} \right)}}$

$ = {{P\left( {{E_1}} \right) - \left[ {P\left( {{E_1} \cap {E_2}} \right) + P\left( {{E_1} \cap {E_3}} \right) - P\left( {{E_1} \cap {E_2} \cap {E_3}} \right)} \right]} \over {P\left( {{E_1}} \right)}}$

$ = {{P\left( {{E_1}} \right) - P\left( {{E_1}} \right).P\left( {{E_2}} \right) - P\left( {{E_1}} \right).P\left( {{E_3}} \right) - 0} \over {P\left( {{E_1}} \right)}}$

$ = 1 - P\left( {{E_2}} \right) - P\left( {{E_3}} \right)$

$ = 1 - P\left( {{E_3}} \right) - P\left( {{E_2}} \right)$

$ = P\left( {E_3^C} \right) - P\left( {{E_2}} \right)$

Let B₁ be the event where Box-I is selected.

And B₂ be the event where Box-II is selected.

P(B₁) = P(B₂) = ${1 \over 2}$

Let E be the event where selected card is non prime.

For B₁ : Prime numbers: {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

For B₂ : Prime numbers: {31, 37, 41, 43, 47}

P(E) = P(B₁) $ \times $ $P\left( {{E \over {{B_1}}}} \right)$ + P(B₂) $ \times $ $P\left( {{E \over {{B_2}}}} \right)$

= ${1 \over 2} \times {{20} \over {30}}$ + ${1 \over 2} \times {{15} \over {20}}$

Required probability :

$P\left( {{{{B_1}} \over E}} \right)$ = ${{P\left( {{B_2}} \right).P\left( {{E \over {{B_1}}}} \right)} \over {P\left( E \right)}}$

= ${{{1 \over 2} \times {{20} \over {30}}} \over {{1 \over 2} \times {{20} \over {30}} + {1 \over 2}{{15} \over {20}}}}$

= ${{{2 \over 3}} \over {{2 \over 3} + {3 \over 4}}}$

= ${8 \over {17}}$

Total ways of distribution = 4¹⁰ = 2²⁰

Number of ways selecting two boxes out of four = ⁴C₂

Then number of ways selecting 5 balls out of 10 = ¹⁰C₅

Then no of ways of distributing 5 balls into two groups of 2 balls and 3 balls = ⁵C₃.2!

Then number of ways to distributing remaining balls into two boxes = 2⁵

Number of ways placing exactly 2 and 3 balls in two of these boxes

= ⁴C₂ $ \times $ ¹⁰C₅ $ \times $ ⁵C₃.2! $ \times $ 2⁵

= ${{{{6.252.10.2.2}^5}} \over {{2^{20}}}}$

= ${{945} \over {{2^{10}}}}$

$\sum\limits_{i = 1}^5 {P(X)} $ = 1

$ \Rightarrow $ K² + 2K + K + 2K + 5K² = 1

$ \Rightarrow $ 6K² + 5K – 1 = 0

$ \Rightarrow $ (6K - 1)(k + 1) = 0

$ \Rightarrow $ K = ${1 \over 6}$ and K = -1(rejected)

$ \therefore $ P(X $ > $ 2)

= K + 2K + 5K²

= ${1 \over 6} + {2 \over 6} + {5 \over {36}}$

= ${{23} \over {36}}$

Possibilities that the second A card appears before the third B card are
=AA + ABA + BAA + ABBA + BBAA + BABA

= ${\left( {{1 \over 2}} \right)^2}$ + ${\left( {{1 \over 2}} \right)^3}$ + ${\left( {{1 \over 2}} \right)^3}$ + ${\left( {{1 \over 2}} \right)^4}$ + ${\left( {{1 \over 2}} \right)^4}$ + ${\left( {{1 \over 2}} \right)^4}$

= ${1 \over 4}$ + ${1 \over 8}$ + ${1 \over 8}$ + ${1 \over {16}}$ + ${1 \over {16}}$ + ${1 \over {16}}$

= ${{11} \over {16}}$

Probability that exactly one of them occurs

P(A) + P(B) – 2P (A $ \cap $ B) = ${2 \over 5}$ .....(1)

Probability that A or B occurs is

P(A) + P(B) – P(A $ \cap $ B) = ${1 \over 2}$ ......(2)

Doing (2) - (1)

P(A $ \cap $ B) = ${1 \over 2} - {2 \over 5}$ = 0.10

Given P(A) = ${1 \over 3}$ and P(B) = ${1 \over 6}$

A and B are independent

So P(A$ \cap $B) = ${1 \over 3} \times {1 \over 6}$ = ${1 \over {18}}$

P(A$ \cup $B) = P(A) + P(B) – P(A $ \cap $ B)

= ${1 \over 3}$ + ${1 \over 6}$ - ${1 \over {18}}$

= ${4 \over 9}$

$P\left( {{A \over {B'}}} \right)$

= ${{P\left( {A \cap B'} \right)} \over {P\left( {B'} \right)}}$

= ${{P\left( A \right) - P\left( {A \cap B} \right)} \over {1 - P\left( B \right)}}$

= ${{{1 \over 3} - {1 \over {18}}} \over {1 - {1 \over 6}}}$ = ${{1 \over 3}}$

Probablity of at most two machines will be out of service = ${\left( {{3 \over 4}} \right)^3}k$

$ \Rightarrow $ ⁵C₀${\left( {{3 \over 4}} \right)^5}$ + ⁵C₁$\left( {{1 \over 4}} \right){\left( {{3 \over 4}} \right)^5}$ + ⁵C₂${\left( {{1 \over 4}} \right)^2}{\left( {{3 \over 4}} \right)^5}$ = ${\left( {{3 \over 4}} \right)^3}k$

$ \Rightarrow $ ${{17} \over 8}{\left( {{3 \over 4}} \right)^3}$ = ${\left( {{3 \over 4}} \right)^3}k$

$ \Rightarrow $ k = ${{17} \over 8}$

Number of ways 3 consecutive heads can appers

(1) HHHT_

(2) _THHH

(3) THHHT

$ \therefore $ Probablity of getting 3 consecutive heads

= ${2 \over {32}}$ + ${2 \over {32}}$ + ${1 \over {32}}$ = ${5 \over {32}}$

Number of ways 4 consecutive heads can appers

(1) HHHHT

(2) THHHH

$ \therefore $ Probablity of getting 4 consecutive heads

= ${1 \over {32}}$ + ${1 \over {32}}$ = ${2 \over {32}}$

Number of ways 5 consecutive heads can appers

(1) HHHHH

$ \therefore $ Probablity of getting 5 consecutive heads

= ${1 \over {32}}$

Now Probablity of getting 0, 1, and 2 consecutive heads

= 1 - $\left( {{5 \over {32}} + {2 \over {32}} + {1 \over {32}}} \right)$ = ${{{24} \over {32}}}$

Now, Expectation

= (-1) $ \times $ ${{{24} \over {32}}}$ + 3 $ \times $ ${{{5} \over {32}}}$ + 4 $ \times $ ${{{2} \over {32}}}$ + 5 $ \times $ ${{{1} \over {32}}}$

= ${1 \over 8}$

When two dice are thrown then sample space will {(1, 1), (2, 2) ....... (6, 6)} contain total 36 elements number of cases.

Then the expectation will be ${6 \over {36}} \times 15 \times {4 \over {36}} \times 12 - {{26} \over {36}} \times 6$

${{90 + 48 - 156} \over {36}} = - {1 \over 2}$ = $ {1 \over 2}$ loss

There are 50 questions in an exam.

So Probability of each question to be correct is p = ${4 \over 5}$

and also probability of each question to be incorrect is q = ${1 \over 5}$

Let Y is the number of correct question in 50 questions, hence required probability is

${\left( {{4 \over 5}} \right)^{50}} + {}^{50}{C_1}\left( {{1 \over 5}} \right){\left( {{4 \over 5}} \right)^{49}}$

= ${\left( {{4 \over 5}} \right)^{49}} + \left( {{4 \over 5} + 10} \right)\left( {{4 \over 5}} \right)$

= ${{54} \over 5}{\left( {{4 \over 5}} \right)^{49}}$

Choosing vertices of a regular hexagon alternate, here A₁, A₃, A₅ or A₂, A₄, A₆ will result in an equilateral triangle.

Hence the required probability = ${2 \over {{}^6{C_3}}}$ = ${1 \over {10}}$

Let number of trials be n and probability of success = p, probability of failure = q

Given np = 8, npq = 4

$ \Rightarrow $ q = ${1 \over 2}$, p = ${1 \over 2}$, n = 16 (as p + q = 1)

p$\left( {x \le 2} \right)$ = ${{{}^{16}{C_0} + {}^{16}{C_1} + {}^{16}{C_2}} \over {{2^{16}}}} = {{1 + 16 + 120} \over {{2^{16}}}} = {{137} \over {{2^{16}}}}$

$1 - {\left( {{1 \over 2}} \right)^n} > {{99} \over {100}}$

${\left( {{1 \over 2}} \right)^n} < {1 \over {100}}$

$ \Rightarrow $ n = 7

A = At least two girls

B = All girls

$P\left( {{B \over A}} \right) = {{P\left( {B \cap A} \right)} \over {P\left( A \right)}}$

$ \Rightarrow {{P(B)} \over {P(A)}} = {{{{\left( {{1 \over 4}} \right)}^2}} \over {1 - {}^4{C_0}{{\left( {{1 \over 2}} \right)}^4} - {}^4{C_1}{{\left( {{1 \over 2}} \right)}^4}}}$

$ \Rightarrow {1 \over {16 - 1 - 4}} = {1 \over {11}}$

Let four persons are A, B, C and D.

Probablity of hitting a target by them,

P(A) = ${1 \over 2}$

P(B) = ${1 \over 3}$

P(C) = ${1 \over 4}$

P(D) = ${1 \over 8}$

Probablity of hitting target atleast once = 1 - Probablity of not hitting by anybody

P(Hit) = 1 - $P\left( {\overline A \cap \overline B \cap \overline C \cap \overline D } \right)$

= 1 - $P\left( {\overline A } \right).P\left( {\overline B } \right).P\left( {\overline C } \right).P\left( {\overline D } \right)$

= 1 - ${1 \over 2}.{2 \over 3}.{3 \over 4}.{7 \over 8}$

= ${{25} \over {32}}$

Probablity of getting head P(H) = ${1 \over 2}$

Probablity of getting tail P(T) = 1 - ${1 \over 2}$ = ${1 \over 2}$

Probability of observing at least one head out of n tosses

= 1 - Probability of observing no head occurs out of n tosses

= 1 - ${\left( {{1 \over 2}} \right)^n}$

According to the question,

1 - ${\left( {{1 \over 2}} \right)^n}$ $ \ge $ ${{90} \over {100}}$

$ \Rightarrow $ ${\left( {{1 \over 2}} \right)^n} \le {1 \over {10}}$

$ \Rightarrow $ ${2^n} \ge 10$

As 2³ = 8

2⁴ = 16

$ \therefore $ Minimum value of n = 4

$P\left( {{A \over B}} \right) = {{P\left( {A \cap B} \right)} \over {P\left( B \right)}}$

As A $ \subset $ B,

then P(A$ \cap $B) = P(A)

$ \therefore $ $P\left( {{A \over B}} \right) = {{P\left( A \right)} \over {P\left( B \right)}}$

As P(B) $ \le $ 1

$ \therefore $ ${{P\left( A \right)} \over {P\left( B \right)}}$ $ \ge $ P(A)

A $ \to $ opted NCC

B $ \to $ opted NSS

$ \therefore $  P (nither A nor B) $=$ ${{10} \over {60}} = z{1 \over 6}$

Expected Gain/Loss =

= w $ \times $ 100 + Lw($-$ 50 + 100)

       + L²w ($-$ 50 $-$ 50 + 100) + L³($-$ 150)

= ${1 \over 3} \times 100 + {2 \over 3}.{1 \over 3}\left( {50} \right) + {\left( {{2 \over 3}} \right)^2}\left( {{1 \over 3}} \right)\left( 0 \right)$

        $ + {\left( {{2 \over 3}} \right)^3}\left( { - 150} \right) = 0$

here w denotes probability that outcome 5 or 6 (w = ${2 \over 6} = {1 \over 3}$)

here L denotes probability that outcome

1,2,3,4 (L = ${4 \over 6}$ = ${2 \over 3}$)

$\underline {} \,\,\,\underline {} \,\,\,\underline {} \,\,\underline 4 \,\,\underline 4 $

${1 \over {{6^2}}}\left( {{{{5^3}} \over {{6^3}}} + {{2{C_1}{{.5}^2}} \over {{6^3}}}} \right) = {{175} \over {{6^5}}}$

We can solve this problem by counting the number of "nice" subsets in the set S = {1, 2, $\ldots$, 20 }, and then dividing that number by the total number of possible subsets of S.

The sum of all elements in S is :

1 + 2 + $\ldots$ + 20 = $\frac{{20 \times 21}}{2}$ = 210

Since a "nice" subset must sum to 203, the elements not in the subset must sum to 210 - 203 = 7.

Now we need to find the ways to make the sum of 7 using the elements of S. The combinations are :

1. 1. 7
2. 2. 1 + 6
3. 3. 2 + 5
4. 4. 3 + 4
5. 5. 1 + 2 + 4
6. 6. 1 + 3 + 3(This doesn't work since 3 is repeated)
7. 7. 2 + 2 + 3(This doesn't work since 2 is repeated)

So, there are 5 "nice" subsets.

Since the set S has 20 elements, there are $2^{20}$ possible subsets (including the empty set and the set itself). The probability of randomly choosing a "nice" subset is therefore :

p (probability of getting white ball) = ${{30} \over {40}}$

q = ${1 \over 4}$ and n = 16

mean = np = 16.${3 \over 4}$ = 12

and standard diviation

= $\sqrt {npq} $ = $\sqrt {16.{3 \over 4}.{1 \over 4}} = \sqrt 3 $

$ \because $ $mean \over standard\,deviation$ = $12\over{\sqrt3}$ = $4{\sqrt3}$

Since sum of two numbers is even so either both are odd or both are even. Hence number of elements in reduced samples space = ⁵C₂ + ⁶C₂

So, required probability = ${{{}^5{C_2}} \over {{}^5{C_2} + {}^6{C_2}}}$ = ${2 \over 5}$

$1 - {}^n{C_0}{\left( {{1 \over 3}} \right)^0}{\left( {{2 \over 3}} \right)^n} > {5 \over 6}$

${1 \over 6} > {\left( {{2 \over 3}} \right)^n}\,\, \Rightarrow \,\,0.1666 > {\left( {{2 \over 3}} \right)^n}$

${n_{\min }} = 5$

$P\left( A \right) = {1 \over 2} \times {{11} \over {36}} + {1 \over 2} \times {2 \over 9} = {{19} \over {72}}$

5 Red and 2 green balls

P(one red ball) = ${5 \over 7}$

P(one green ball) = ${2 \over 7}$

Case I :

If drawn ball is green than a red ball is added

$\left( {\matrix{ {6{\mathop{\rm Re}\nolimits} d} \cr {1\,Green} \cr } } \right)$ P (red ball) = ${6 \over 7}$

Case II :

If drawn ball is red than a green ball is added

$\left( {\matrix{ {4{\mathop{\rm Re}\nolimits} d} \cr {3\,Green} \cr } } \right)$ P (red ball) = ${4 \over 7}$

P (2^nd red ball) = ${5 \over 7}$ $ \times {4 \over 7} + {2 \over 7} \times {6 \over 7}$ = ${{32} \over {49}}$

P (X = 1) means out of two drawn cards one card is ace.

and P(X = 2) means both the drawn cards are ace.

$ \therefore $  P(X = 1) = first card is ace or 2nd card is ace.

= A _ $+$ _ A

= ${4 \over {52}} \times {{48} \over {52}} + {{48} \over {52}} \times {4 \over {52}}$

= $2 \times {4 \over {52}} \times {{48} \over {52}}$

P(X = 2) = First and second both cards arc ace.

= A A

= ${4 \over {52}} \times {4 \over {52}}$

$ \therefore $  P(X = 1) + P(X = 2)

= $2 \times {4 \over {52}} \times {{48} \over {52}} + {4 \over {52}} \times {4 \over {52}}$

= ${{25} \over {169}}$

Here, $P\left( {\overline A \cap \overline B \left| C \right.} \right) = {{P\left( {\overline A \cap \overline B \cap C} \right)} \over {P\left( C \right)}}$

= ${{P\left[ {\left( {\overline {A \cup B} } \right) \cap C} \right]} \over {P\left( C \right)}}$

= ${{P\left[ {C - \left( {A \cup B} \right)} \right]} \over {P\left( C \right)}}$

= ${{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right) + P\left( {A \cap B \cap C} \right)} \over {P\left( C \right)}}$

= ${{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right)} \over {P\left( C \right)}}$ ($ \because $$\left. {P\left( {A \cap B \cap C} \right) = 0} \right)$

= ${{P\left( C \right) - P\left( A \right).P(C) - P\left( B \right).P(C)} \over {P\left( C \right)}}$

[$ \because $ A, B and C are independent events]

= 1 - P(A) - P(B)

= $P\left( {\overline A } \right)$ - P(B) or $P\left( {\overline B } \right)$ - P(A)

Let the number of children in each family be x.

Thus the total number of children in both the families are 2x

Now, it is given that 3 tickets are distributed amongst the children of these two families.

Thus, the probability that all the three tickets go to the children in family B

= ${{{}^x{C_3}} \over {{}^{2x}{C_3}}}$ = ${1 \over {12}}$

$ \Rightarrow $ $\,\,\,$ ${{x\left( {x - 1} \right)\left( {x - 2} \right)} \over {2x\left( {2x - 1} \right)\left( {2x - 2} \right)}}$ = ${1 \over {12}}$

$ \Rightarrow $  ${{\left( {x - 2} \right)} \over {\left( {2x - 1} \right)}}$ = ${1 \over 6}$

Thus, the number of children in each family is 5.

If we follow path 1, then probability of getting 1st ball black $ = {6 \over {10}}$ and probability of getting 2nd ball red when there is 4 R and 8 B balls = ${4 \over {12}}$.

So, the probability of getting 1st ball black and 2nd ball red = ${6 \over {10}} \times {4 \over {12}}$.

If we follow path 2, then the probability of getting 1st ball red $ = {4 \over {10}}$ and probability of getting 2nd ball red when in the bag there is 6 red and 6 black balls = ${6 \over {12}}$

$\therefore\,\,\,$ Probability of getting 2nd ball as red

$ = {6 \over {10}} \times {4 \over {12}} + {4 \over {10}} \times {6 \over {12}}$

$ = {1 \over 5} + {1 \over 5}$

$ = {2 \over 5}$

P(X getting head) = p

$ \therefore $ P(X getting tail) = 1 - p

P(Y getting head) = P(Y getting tail) = ${1 \over 2}$

P(X wins) = p + (1 - p)${1 \over 2}$p + (1 - p)${1 \over 2}$(1 - p)${1 \over 2}$p + ...

= ${p \over {1 - \left( {{{1 - p} \over 2}} \right)}}$

= ${{2p} \over {1 + p}}$

P(Y win) = (1 - p)${1 \over 2}$ + (1 - p)${1 \over 2}$(1 - p)${1 \over 2}$ + ...

= $\left( {{{1 - p} \over 2}} \right).{p \over {1 - \left( {{{1 - p} \over 2}} \right)}} = {{1 - p} \over {1 + p}}$

According to question,

P(X wins) = P(Y wins)

$ \therefore $ ${{2p} \over {1 + p}}$ = ${{1 - p} \over {1 + p}}$

$ \Rightarrow $ 3p = 1

$ \Rightarrow $ p = ${1 \over 3}$

Probability of drawing a white ball and then a red ball

from bag B is given by ${{{}^4{C_1} \times {}^2{C_1}} \over {{}^9{C_2}}}$ = ${2 \over 9}$

Probability of drawing a white ball and then a red ball

from bag A is given by ${{{}^2{C_1} \times {}^3{C_1}} \over {{}^7{C_2}}}$ = ${2 \over 7}$

Hence, the probability of drawing a white ball and then

a red ball from bag B = ${{{2 \over 9}} \over {{2 \over 7} + {2 \over 9}}}$ = ${{2 \times 7} \over {18 + 14}}$ = ${7 \over {16}}$

Let P(E) = x and P(F) = y

Now, $P(E \cap F) = {1 \over {12}}$

$ \Rightarrow P(E)P(F) = {1 \over {12}}$

$ \Rightarrow xy = {1 \over {12}}$

Also, $P(E' \cap F') = {1 \over 2}$

$ \Rightarrow (1 - P(E))(1 - P(F)) = {1 \over 2}$

$ \Rightarrow (1 - x)(1 - y) = {1 \over 2}$

$ \Rightarrow 1 - x - y + xy = {1 \over 2}$

$ \Rightarrow x + y = 1 + xy - {1 \over 2}$

$ \Rightarrow x + y = 1 + {1 \over {12}} - {1 \over 2}$

$ \Rightarrow x + y = {1 \over 2} + {1 \over {12}} = {7 \over {12}}$ ........ (1)

Now,

${(x - y)^2} = {(x + y)^2} - 4xy$

$ \Rightarrow {(x - y)^2} = {{49} \over {144}} - {1 \over 3} \Rightarrow {{49 - 98} \over {144}} \Rightarrow {1 \over {144}}$

$ \Rightarrow (x - y) = {1 \over {12}} \Rightarrow x - y = {1 \over 2}$ ........ (2)

From Eqs. (1) and (2), we get

$(x + y)(x - y) = {7 \over {12}} + {1 \over {12}}$

$ \Rightarrow x = {4 \over {12}};y = {3 \over {12}}$

$ \Rightarrow {x \over y} = {4 \over 3} = {{P(E)} \over {P(F)}}$

The number of ways to form a committee having at least one woman is

$ = {}^5{C_1} \times {}^{10}{C_3} + {}^5{C_2} \times {}^{10}{C_2} + {}^5{C_3} \times {}^{10}{C_1} + {}^5{C_4}$

$ = {{5!} \over {4!}} \times {{10!} \over {7! \times 3!}} + {{5!} \over {2!3!}} \times {{10!} \over {8!2!}} + {{5!} \over {3!2!}} \times {{10!} \over {9!1!}} + {{5!} \over {4!}}$

$ = 5 \times {{10 \times 9 \times 8} \over {3 \times 2}} + {{5 \times 4} \over {2 \times 1}} \times {{10 \times 9} \over 2} + {{5 \times 4} \over {2 \times 1}} \times 10 + 5$

$ = 600 + 450 + 100 + 5 = 1155$

The number of ways to form a committee having more women than men is

${}^5{C_2} \times {}^{10}{C_2} + {}^5{C_4} = {{5!} \over {6!2!}} \times {{10!} \over {9!}} + {{5!} \over {4!}}$

$ = 10 \times 10 + 5 = 105$

Therefore, the probability for the committees to have more women than men is

${{105} \over {1155}} = {1 \over {11}}$

An unbiased coin is tossed 8 times which is same as 8 coins tossed 1 times.

$\therefore\,\,\,$ Possible no. of out come = 2⁸

$\therefore\,\,\,$ Sample space = 2⁸

Here in this condition, all head or all tail out come is not acceptable.

No. of times all head can occur

(H H H H H H H H) = 1

$\therefore\,\,\,$ Probability (all head) = ${1 \over {{2^8}}}$ = ${1 \over {256}}$

No. of times all tail can occur

(T T T T T T T T) = 1

$\therefore\,\,\,$ Probability (all tail) = ${1 \over {{2^8}}}$ = ${{1 \over {256}}}$

$\therefore\,\,\,$ Required probability

= 1 $-$ (P (All head) + P (All tail))

= 1 $-$ ( ${{1 \over {256}}}$ + ${{1 \over {256}}}$)

= 1 $-$ ${{1 \over {128}}}$

= ${{127} \over {128}}$

We have the following probabilities:

$\bullet$ The probability that the target is hit by the person P is ${3 \over 4}$.

$\bullet$ The probability that the target is not hit by the person P is $1 - {3 \over 4} = {1 \over 4}$.

$\bullet$ The probability that the target is hit by the person Q is ${1 \over 2}$.

$\bullet$ The probability that the target is not hit by the person Q is $1 - {1 \over 2} = {1 \over 2}$.

$\bullet$ The probability that the target is hit by the person R is ${5 \over 8}$.

$\bullet$ The probability that the target is not hit by the person R is $1 - {5 \over 8} = {3 \over 8}$.

Here, we have used the fact that if the probability of occurrence of an event is p, then the probability of non-occurrence of an event is $q = 1 - p$.

Therefore, the probability that the target is hit by P or Q and not by R is

(Probability that the target is hit by P and not by Q and R) + (Probability that the target is hit by Q and not by P and R) + (Probability that the target is hit by both P and Q and not by R)

$ = \left( {{3 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right) + \left( {{1 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right) + \left( {{3 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right)$

$ = {9 \over {64}} + {3 \over {64}} + {9 \over {64}} = {{9 + 3 + 9} \over {64}} = {{21} \over {64}}$

We can apply binomial probability distribution

n = 10

p = Probability of drawing a green ball = ${{15} \over {25}}$ = ${3 \over 5}$

Also q = 1 - ${3 \over 5}$ = ${2 \over 5}$

Variance = npq

= $10 \times {3 \over 5} \times {2 \over 5}$ = ${{12} \over 5}$

Let A = {0, 1, 2, 3, 4, ......., 10}

Total number of ways of selecting 2 different numbers from A is

n (S) = ¹¹C₂ = 55, where 'S' denotes sample space

Let E be the given event

$ \therefore $ E = {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}

$ \Rightarrow $ n (E) = 6

$ \therefore $ P(E) = ${{n\left( E \right)} \over {n\left( S \right)}}$ = ${6 \over {55}}$

Given, P (A $ \cap $ B $ \cap $ C) = ${1 \over {16}}$

P (exactly one of A or B occurs)

= P(A) + P (B) – 2P (A $ \cap $ B) = ${1 \over 4}$ .....(1)

P (Exactly one of B or C occurs)

= P(B) + P (C) – 2P (B $ \cap $ C) = ${1 \over 4}$ .....(2)

P (Exactly one of C or A occurs)

= P(C) + P(A) – 2P (C $ \cap $ A) = ${1 \over 4}$ .....(3)

Adding (1), (2) and (3),we get

2[ P(A) + P(B) + P (C) - P (A $ \cap $ B)

- P (B $ \cap $ C) - P (C $ \cap $ A)] = ${3 \over 4}$

$ \Rightarrow $ P(A) + P(B) + P (C) - P (A $ \cap $ B)

- P (B $ \cap $ C) - P (C $ \cap $ A) = ${3 \over 8}$

$ \therefore $ P(atleast one event occurs)

= P (A $ \cup $ B $ \cup $ C)

= P(A) + P(B) + P (C) - P (A $ \cap $ B)

- P (B $ \cap $ C) - P (C $ \cap $ A) + P (A $ \cap $ B $ \cap $ C)

= ${3 \over 8} + {1 \over {16}}$ = ${7 \over {16}}$

Probability of fail, P(F) = p

$ \therefore $   Probability of success, P(S) = 2p

We know,

p + 2Pp = 1

$ \Rightarrow $   p = ${1 \over 3}$

$ \therefore $   Now, probability of at least 5 success in 6 trials,

P(x $ \ge $ 5)

= P(x = 5) + P (x = 6)

= ⁶C₅ ${\left( {{2 \over 3}} \right)^5}{\left( {{1 \over 3}} \right)^1}{ + ^6}{C_6}{\left( {{2 \over 3}} \right)^6}{\left( {{1 \over 3}} \right)^o}$

= ${\left( {{2 \over 3}} \right)^5}\left( {{6 \over 3} + {2 \over 3}} \right)$

= ${{256} \over {729}}$

$P\left( {{A \over {A' \cup B'}}} \right)$

$ = {{P\left[ {A \cap \left( {A' \cup B'} \right)} \right]} \over {P\left( {A' \cap B'} \right)}}$

$ = {{P\left[ {\left( {A \cap A'} \right) \cup \left( {A \cap B'} \right)} \right]} \over {P\left( {A \cap B} \right)'}}$

$\left[ \, \right.$As   $\left. {\left( {A \cap B} \right)' = A' \cap B'} \right]$

$ = {{P\left( {A \cap B'} \right)} \over {P\left( {A \cap B} \right)'}}$

As   $A \cap A' = \phi $

$ \therefore $   $\phi \cup \left( {A \cap B'} \right) = A \cap B'$

$ = {{P\left( A \right) - P\left( {A \cap B} \right)} \over {1 - P\left( {A \cap B} \right)}}$

$\left[ \, \right.$As   $A \cap B' = A - \left( {A \cap B} \right)$

and   $\left. {\left( {A \cap B} \right)' = 1 - \left( {A \cap B} \right)} \right]$

Given  $P\left( A \right) = {2 \over 5}$

and   $P\left( {A \cap B} \right) = {3 \over {20}}$

$ = {{{2 \over 5} - {3 \over {20}}} \over {1 - {3 \over {20}}}}$

$ = {{{{8 - 3} \over {20}}} \over {{{17} \over {20}}}}$

$ = {5 \over {17}}$

Total possible outcome with two six faced dice = 6² = 36

When dice A shows up 4, the possible cases are
E₁ = { (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) } = 6 cases
$\therefore$ $P\left( {{E_1}} \right) = {6 \over {36}} = {1 \over 6}$

When dice B shows up 2, the possible cases are
E₂ = { (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) } = 6 cases

$P\left( {{E_2}} \right) = {6 \over {36}} = {1 \over 6}$

${E_1} \cap {E_2}$ = Common in both in E₁ and E₂ = { (4, 2) }

$P\left( {{E_1} \cap {E_2}} \right) = {1 \over {36}}$

And $P\left( {{E_1}} \right)$.$P\left( {{E_2}} \right)$ = ${1 \over 6}$.${1 \over 6}$ = ${1 \over 36}$

$\therefore$$P\left( {{E_1} \cap {E_2}} \right)$ = $P\left( {{E_1}} \right)$.$P\left( {{E_2}} \right)$

$\therefore$ E₁ and E₂ are independent.

${E_3}$ = [ (1, 2), (1, 4), (1, 6),
           (2, 1), (2, 3), (2, 5),
           (3, 2), (3, 4), (3, 6),
           (4, 1), (4, 3), (4, 5),
           (5, 2), (5, 4), (5, 6),
           (6, 1), (6, 3), (6, 5) ] = 18 cases

${E_1} \cap {E_3}$ = { (4, 1) (4, 3) (4, 5) } = 3 cases

$\therefore$ $P\left( {{E_1} \cap {E_3}} \right) = {3 \over {36}} = {1 \over {12}}$ = ${1 \over 6} \times {1 \over 2}$ = $P\left( {{E_1}} \right)$ $ \times $ $P\left( {{E_3}} \right)$

$\therefore$ E₁ and E₃ are independent.

${E_2} \cap {E_3}$ = { (1, 2) (3, 2) (5, 2) } = 3 cases

$\therefore$ $P\left( {{E_2} \cap {E_3}} \right) = {3 \over {36}} = {1 \over {12}}$ = ${1 \over 6} \times {1 \over 2}$ = $P\left( {{E_2}} \right)$ $ \times $ $P\left( {{E_3}} \right)$

$\therefore$ E₂ and E₃ are independent.

${E_1} \cap {E_2} \cap {E_3}$ = 0

$\therefore$ $P\left( {{E_1} \cap {E_2} \cap {E_3}} \right)$ = 0

$P\left( {{E_1}} \right) \times $ $P\left( {{E_2}} \right) \times $$P\left( {{E_3}} \right)$ = ${1 \over 6}$ $ \times $ ${1 \over 6}$ $ \times $ ${1 \over 2}$ = ${1 \over {72}}$

$\therefore$ ${E_1},$ ${E_2}$ and ${E_3}$ are not independent.

$\therefore$ Option (D) is correct.