An irregular six-faced die is thrown. The pro | vedclass

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(B) Let $p$ be the probability of getting an even number and $q = 1 - p$ be the probability of getting an odd number.Let the random variable $X \sim B

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(B) Let $p$ be the probability of getting an even number and $q = 1 - p$ be the probability of getting an odd number.Let the random variable $X \sim B(n, p)$ where $n = 5$.Given that $P(X = 3) = 2 P(X = 2)$.Using the binomial probability formula $P(X = k) = { }^n C_k p^k q^{n-k}$:${ }^5 C_3 p^3 q^2 = 2 	imes { }^5 C_2 p^2 q^3$.Since ${ }^5 C_3 = 10$ and ${ }^5 C_2 = 10$,we have $10 p^3 q^2 = 20 p^2 q^3$.Dividing both sides by $10 p^2 q^2$ (assuming $p, q 
eq 0$),we get $p = 2q$.Since $p + q = 1$,substituting $p = 2q$ gives $3q = 1$,so $q = \frac{1}{3}$ and $p = \frac{2}{3}$.The probability of getting no even number in $5$ throws is $P(X = 0) = { }^5 C_0 p^0 q^5 = q^5 = (\frac{1}{3})^5 = \frac{1}{243}$.In $6804$ sets of $5$ throws,the expected number of times to get no even number is $6804 	imes \frac{1}{243} = 28$.

An irregular six-faced die is thrown. The probability that in $5$ throws it will give $3$ even numbers is twice the probability that it will give $2$ even numbers. The number of times,in $6804$ sets of $5$ throws,you expect to get no even number is:

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