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Exercise 11.5 · Q9

Q.In a binomial distribution consisting of 55 independent trials, the probabilities of 11 and 22 successes are 0.40960.4096 and 0.20480.2048 respectively. Find the mean and variance of the random variable.

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Dividing the two given binomial probabilities cancels (5x)\binom5x-type factors down to a simple ratio in p,qp,q, which pins down pp directly; mean and variance then follow from n=5n=5.

Step 1. Write both probabilities. P(X=1)=(51)pq4=5pq4=0.4096P(X=1)=\dbinom51pq^4=5pq^4=0.4096 and P(X=2)=(52)p2q3=10p2q3=0.2048P(X=2)=\dbinom52p^2q^3=10p^2q^3=0.2048.

Step 2. Take the ratio to eliminate common factors. P(X=2)P(X=1)=10p2q35pq4=2pq=0.20480.4096=0.5\dfrac{P(X=2)}{P(X=1)}=\dfrac{10p^2q^3}{5pq^4}=\dfrac{2p}{q}=\dfrac{0.2048}{0.4096}=0.5.

Step 3. Solve for pp in terms of qq, then use p+q=1p+q=1. 2pq=0.5⇒p=0.25q\dfrac{2p}q=0.5\Rightarrow p=0.25q. Substituting into p+q=1p+q=1: 0.25q+q=1⇒1.25q=1⇒q=0.8, p=0.20.25q+q=1\Rightarrow1.25q=1\Rightarrow q=0.8,\ p=0.2. …

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