Probability

To get a sum of 9 ,possible dice combinations are

$ \begin{array}{ll} (1,3,6) & -6 \text { ways } \\ (1,3,5) & -5 \text { ways } \\ (1,4,4) & -3 \text { ways } \\ (2,3,5) & -3 \text { ways } \\ (2,3,4) & -6 \text { ways } \\ (3,3,3) & -1 \text { way } \end{array} $

Total ways $=6+6+3+3+6+1=25$

Total possible outcomes when rolling three dice $=6 \times 6 \times 6=216$.

Number of favourable outcomes $=25$

∴ Probability of getting a sum of 9 in a single throw of three fair dice $=\frac{25}{216}$

We know that

Mean, $\mu=n p=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Variance, $\sigma^2=n p(1-p)=6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Dividing Eq. (ii) by Eq. (i),

$ \begin{gathered} (1-p)=\frac{6}{10}=0.6 \\ p=0.4 \end{gathered} $

Substituting $p=0.4 \mathrm{in}$ Eq. (i),

$ \begin{aligned} n \times 0.4 & =10 \\ n & =25 \end{aligned} $

The median of a binomial distribution is typically approximated by

$ M \approx[n p] $

Substituting values

$ M=[25 \times 0.4]=[10] $

Thus, the median is 10 .

Total number of ways in which four digit number can be found $=4!=24$

So, exhaustive number of cases $=24$

Now, the number is divisible by 5 if at units place we have 5.

∴ Unit place can be filled in one way the remaining 3 places can be filled with other digits in 3! ways.

So, the number of ways of forming the number divisible by 5 is $3!=6$

So, the required probability $=\frac{6}{24}=\frac{1}{4}$

The probability of rolling a total of 10 with three fair dice can be calculated as follows.

First, identify the total number of possible outcomes when rolling three dice:

$ \text{Total outcomes} = 6 \times 6 \times 6 = 216 $

Next, find the number of successful outcomes where the sum of the dice equals 10. Consider each possible combination:

Combination: (1, 3, 6), (1, 6, 3), (3, 1, 6), etc. (permutations):

Number of outcomes: $3! = 6$

Combination: (2, 4, 4), (4, 4, 2), (4, 2, 4) (permutations of two identical numbers):

Number of outcomes: $\frac{3!}{2!} = 3$

Combination: (2, 3, 5), (3, 2, 5), (5, 3, 2), etc. (permutations):

Number of outcomes: $3! = 6$

Combination: (3, 3, 4), (3, 4, 3), (4, 3, 3) (permutations of two identical numbers):

Number of outcomes: $\frac{3!}{2!} = 3$

Combination: (2, 2, 6), (2, 6, 2), (6, 2, 2) (permutations of two identical numbers):

Number of outcomes: $\frac{3!}{2!} = 3$

Combination: (3, 5, 2), (5, 3, 2), (2, 3, 5), etc. (permutations):

Number of outcomes: $3! = 6$

The total number of favorable outcomes for rolling a sum of 10 is:

$ 6 + 3 + 6 + 3 + 3 + 6 = 27 $

Finally, calculate the probability:

$ \text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{27}{216} = \frac{1}{8} $

Mean $ =n p=4 $

$ \begin{aligned} \text { Variance } & =n p q=3 \\ \Rightarrow \quad q & =\frac{3}{4} \end{aligned} $

and $ \quad p=1-q=1-\frac{3}{4}=\frac{1}{4} $

$ \Rightarrow \quad n=16 $

Now, Mode of Binomial distribution is given by $ (n+1) P $

Case IIf $ (n+1) P=\operatorname{Integer}(I) $, then Mode $ =\{I, I-1\} $

Case II If $ (n+1) P \neq \operatorname{Integer}(I+f) $, then Mode $ =\{I\} $

$ \begin{aligned} \therefore(n+1) P & =(16+1) \frac{1}{4} \\ & =\frac{17}{4}=4.25 \\ & =4+0.25 \\ \Rightarrow \text { Mode } & =4 \end{aligned} $

Let $P$ be the probability of getting an even number. Then, by hypothesis the probability of getting an odd number be $3 P$.

Since, event of getting an odd number and an even number are mutually exclusive and exhaustive.

$\therefore \quad 3 P+P=1 \Rightarrow P=\frac{1}{4}$

Thus, the probability of getting an odd number in a single throw is $\frac{3}{4}$ and that of an even number is $\frac{1}{4}$. If the die is thrown twice, then the sum of the numbers in two throws is odd if one number is odd and another is even.

$\therefore$ Required probability

$=\frac{1}{4} \times \frac{3}{4}+\frac{3}{4} \times \frac{1}{4}=\frac{6}{16}=\frac{3}{8}$

Let P be the probability of getting an even number. Then by hypothesis the probability of getting an odd number is 3P. Since the events of getting an even number and an odd number are mutually exclusive and exhaustive.

$\therefore$ P + 3P = 1 $\Rightarrow$ P = 1/4

Thus, the probability of getting an odd number in a single throw is $\frac{3}{4}$ and that of an even number is $\frac{1}{4}$. If the die is thrown twice, then the sum of the numbers in two throws is even if both the numbers are even or both are odd.

$\therefore$ Required probability = $\frac{3}{4}$ $\times$ $\frac{3}{4}$ + $\frac{1}{4}$ $\times$ $\frac{1}{4}$ = $\frac{10}{16}$ = $\frac{5}{8}$

Let E₁ = event of getting both red cards

E₂ = event of getting both kings

and E₁ $\cap$ E₂ = event of getting 2 kings of red cards

$\therefore$ $P({E_1}) = {{{}^{26}{C_2}} \over {{}^{52}{C_2}}} = {{325} \over {1326}}$, $P({E_2}) = {{{}^4{C_2}} \over {{}^{52}{C_2}}} = {6 \over {1326}}$

and $P({E_1} \cap {E_2}) = {{{}^2{C_2}} \over {{}^{52}{C_2}}} = {1 \over {1326}}$

$\therefore$ P (both red or both kings) = $P({E_1} \cap {E_2}) = P({E_1}) + P({E_2}) - P({E_1} \cap {E_2})$

$ = {{325} \over {1326}} + {6 \over {1326}} - {1 \over {1326}} = {{330} \over {1326}} = {{55} \over {221}}$

Since, A and B are independent events.

$\therefore$ $P(A/B) = P(A) = {1 \over 2} \Rightarrow P\left( {{A \over {A \cup B}}} \right) = {{P[A \cap (A \cup B)]} \over {P(A \cup B)}}$

$ \Rightarrow P\left[ {{A \over {A \cup B}}} \right] = {{P(A)} \over {P(A \cup B)}} = {{1/2} \over {{1 \over 2} + {1 \over 5} - {1 \over {10}}}}$

$ = {{1/2} \over {{1 \over 2} + {1 \over 5} - {1 \over {10}}}} = {{1/2} \over {{6 \over {10}}}} = {5 \over 6}$

Similarly, $P\left( {{{A \cap B} \over {A' \cup B'}}} \right) = P\left( {{{A \cap B} \over {(A \cup B)'}}} \right) = 0$

Let E₁ = coin shows head, E₂ = coin shows tail, A = drawn ball is blue

$P({E_1}) = {1 \over 2} = P({E_2})$

P(A/E₁) = Probability of drawing a blue ball from bag I = 2/7

$P\left( {{A \over {{E_2}}}} \right)$ = Probability of drawing a blue ball from bag II = ${6 \over 8}$

By Baye's theorem, we have

$P\left( {{{{E_1}} \over A}} \right) = {{P({E_1})P\left( {{A \over {{E_1}}}} \right)} \over {P({E_1})P\left( {{A \over {{E_1}}}} \right) + P({E_2})P\left( {{A \over {{E_2}}}} \right)}}$

$ = {{{1 \over 2} \times {2 \over 7}} \over {{1 \over 2} \times {2 \over 7} + {1 \over 2} \times {6 \over 8}}} = {{{2 \over 7}} \over {{2 \over 7} + {6 \over 8}}} = {2 \over 7} \times {{56} \over {58}} = {8 \over {29}}$