Let Xbe the number of heads obtained in a sequence of 10 independent tosses of a fair coin. The fair coin is tossed again Xnumber of times independently, and let Ybe the number of heads obtained in these Xnumber of tosses. Then E(X+ 2Y) equals __________Correct answer is '10'. Can you explain this answer?

Question

Answer

Solution:

Calculation of E(X)

- Let p be the probability of getting a head in a single toss of a fair coin, which is 1/2.
- Let X be the number of heads obtained in a sequence of 10 independent tosses of a fair coin.
- X follows a binomial distribution with parameters n = 10 and p = 1/2.
- The expected value of a binomial distribution is given by E(X) = np, which in this case is 10*(1/2) = 5.

Calculation of E(Y)

- Let Y be the number of heads obtained in X number of tosses of a fair coin.
- Y follows a binomial distribution with parameters n = X and p = 1/2.
- The expected value of a binomial distribution is given by E(Y) = np, where n = X and p = 1/2.
- Substituting np = E(X) = 5, we get E(Y) = 5.

Calculation of E(X+2Y)

- We need to find E(X+2Y), which can be rewritten as E(X) + 2E(Y).
- Substituting the values of E(X) and E(Y), we get E(X+2Y) = 5 + 2*5 = 15.
- Therefore, E(X 2Y) = E(X+2Y) - E(X) = 15 - 5 = 10.

Final Answer:

- E(X 2Y) = 10.

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