Probability

The required probability

$ = {\text{Area of Region PQCAP} \over \text{Area of Region ABCA}}$

$ = {{{1 \over 2} \times 8 \times 6 - {1 \over 2} \times 2 \times 4} \over {{1 \over 2} \times 8 \times 6}}$

$ = {5 \over 6}$

Given P(X = 3) = 5P(X = 4) and n = 7

$ \Rightarrow {}^7{C_3}{p^3}{q^4} = 5\,.\, \Rightarrow {}^7{C_4}{p^4}{q^3}$

$ \Rightarrow q = 5p$ and also $p + q = 1$

$ \Rightarrow p = {1 \over 6}$ and $q = {5 \over 6}$

Mean $ = {7 \over 6}$ and variance $ = {{35} \over {36}}$

Mean + Variance $ = {7 \over 6} + {{35} \over {36}} = {{77} \over {36}}$

Coin is tossed 5 times, so n = 5

Let, p = probability of getting heads

q = probability of getting tails.

$\therefore$ p + q = 1 ...... (1)

$\therefore$ Probability of getting 4 heads

= ⁵C₄ . p⁴ . q

And probability of getting 5 heads

= ⁵C₅ . p⁵

Given, ⁵C₄ . p⁴ . q = ⁵C₅ . p⁵

$\Rightarrow$ 5q = p ....... (2)

From equation (1) and (2), we get,

5q + q = 1

$\Rightarrow$ 6q = 1

$\Rightarrow$ q = ${1 \over 6}$

$\therefore$ p = 1 $-$ ${1 \over 6}$ = ${5 \over 6}$

Now, probability of getting atmost two heads

= p (x = 0) + p (x = 1) + p (x = 2)

p (x = 0) = Getting zero head in 5 trials

= ⁵C₀ . p⁰ . q⁵

p (x = 1) = Getting one head in 5 trials

= ⁵C₁ . p¹ . q⁴

p (x = 2) = Getting two heads in 5 trials

= ⁵C₂ . p² . q³

= ⁵C₀ . q⁵ + ⁵C₁ . pq⁴ + ⁵C₂ . p²q³

$ = {\left( {{1 \over 6}} \right)^5} + 5\,.\,{5 \over 6}\,.\,{\left( {{1 \over 6}} \right)^4} + 10\,.\,{\left( {{5 \over 6}} \right)^2}\,.\,{\left( {{1 \over 6}} \right)^3}$

$ = {{1 + 25 + 250} \over {{6^5}}} = {{276} \over {{6^5}}}$ = ${{46} \over {{6^4}}}$

There are only two ways to get sum 48, which are (32, 8, 8) and (16, 16, 16)

So, required probability

$ = 3\left( {{2 \over {32}}\,.\,{1 \over 8}\,.\,{1 \over 8}} \right) + \left( {{1 \over {16}}\,.\,{1 \over {16}}\,.\,{1 \over {16}}} \right)$

$ = {3 \over {{2^{10}}}} + {1 \over {{2^{12}}}}$

$ = {{13} \over {{2^{12}}}}$

$P\left( {{{{E_1}} \over {{E_2}}}} \right) = {1 \over 2} \Rightarrow {{P({E_1} \cap {E_2})} \over {P({E_2})}} = {1 \over 2}$

$P\left( {{{{E_2}} \over {{E_1}}}} \right) = {3 \over 4} \Rightarrow {{P({E_2} \cap {E_1})} \over {P({E_1})}} = {3 \over 4}$

$P({E_1} \cap {E_2}) = {1 \over 8}$

$P({E_2}) = {1 \over 4},\,P({E_1}) = {1 \over 6}$

(A) $P({E_1} \cap {E_2}) = {1 \over 8}$ and $P({E_1})\,.\,P({E_2}) = {1 \over {24}}$

$ \Rightarrow P({E_1} \cap {E_2}) \ne P({E_1})\,.\,P({E_2})$

(B) $P(E{'_1} \cap E{'_2}) = 1 - P({E_1} \cup {E_2})$

$ = 1 - \left[ {{1 \over 4} + {1 \over 6} - {1 \over 8}} \right] = {{17} \over {24}}$

$P(E{'_1}) = {3 \over 4} \Rightarrow P(E{'_1})P({E_2}) = {3 \over {24}}$

$ \Rightarrow P(E{'_1} \cap E{'_2}) \ne P(E{'_1})\,.\,P({E_2})$

(C) $P({E_1} \cap E{'_2}) = P({E_1}) - P({E_1} \cap {E_2})$

$ = {1 \over 6} - {1 \over 8} = {1 \over {24}}$

$P({E_1})\,.\,P({E_2}) = {1 \over {24}}$

$ \Rightarrow P({E_1} \cap E{'_2}) = P({E_1})\,.\,P({E_2})$

(D) $P(E{'_1} \cap {E_2}) = P({E_2}) - P({E_1} \cap {E_2})$

$ = {1 \over 4} - {1 \over 8} = {1 \over 8}$

$P({E_1})P({E_2}) = {1 \over {24}}$

$ \Rightarrow P(E{'_1} \cap {E_2}) \ne P({E_1})\,.\,P({E_2})$

$\because$ x is a random variable

$\therefore$ $k + 2k + 4k + 6k + 8k = 1$

$\therefore$ $k = {1 \over {21}}$

Now, $P(1 < x < 4\,|\,x \le 2) = {{4k} \over {7k}} = {4 \over 7}$

Let x be the number of heads obtained by A, and y be the number of heads obtained by B.

Note that x and y are binomial variable with parameters n = 3 and p = ${1 \over 2}$

$\therefore$ Probability that both A and B obtained the same number of heads is

$ = P(x = 0)\,.\,P(y = 0) + P(x = 1)\,.\,P(y = 1) + P(x = 2)\,.\,P(y = 2) + P(x = 3)\,.\,P(y = 3)$

$ = {\left[ {{3_{{C_0}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2} + {\left[ {{3_{{C_1}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2} + {\left[ {{3_{{C_2}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2} + {\left[ {{3_{{C_3}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2}$

$ = {\left( {{1 \over 2}} \right)^6}\left[ {1 + 9 + 9 + 1} \right]$

$ = {{20} \over {64}}$

$ = {5 \over {16}}$