Question

1. Roll an even dice and observe the number N on the uppermost face. Then toss a fair coin N times and observe X, the total number of heads that appear in N tosses. (i) Write down the conditional probability mass function pX|N(·|3). (ii) What is P(X = 5)? (iii) What is E(X)?

Answer 1

$a! = a(a-1)(a-2)...1 \\\\ \binom{a}{b} = \frac{a!}{b!(a-b)!} \\\\ $ For a binomial distributed variable $ X $ with \\ sample size$(n) $ and probability of success$(p) \\\\ P(x = k) = \binom{n}{k} * p^{k} * (1-p)^{n-k} \\\\ --------------------------------$

$$ On a dice, there are 6 outcomes : $ \{1,2,3,4,5,6\} \\\\ $ On an even dice, all the 6 outcomes are equally likely $ \\\\ $ So, probability of any outcome on an even dice $ = \frac{1}{6} \\\\ $ On a coin, there are 2 outcomes : $ \{Head,Tail\} \\\\ $ On a fair coin, both the outcomes are equally likely $ \\\\ $ So, probability of any outcome on a fair coin $ = \frac{1}{2}$

$$ An even dice is rolled and then. a fair coin is tossed the same number of times as the outcome on the dice $ \\\\ $ The outcome on the dice is the number on the uppermost face after the dice is rolled $ \\\\ $ The outcome on the dice is represented by random variable $ N , $ so the coin is tossed $ N $ times, and the total number of heads on those $ N $ tosses are represented by the random variable $ X$

$i) \\\\ $ For $ n $ tosses of a fair coin, number of heads $ y $ are a binomial distributed variable with probability of success$(p) = \frac{1}{2} \\\\ $ So, probability of $ k $ heads on $ n $ tosses $ \\\\ = P(y = k) = \binom{n}{k} * \Big(\frac{1}{2}\Big)^{k} * \Big(1 - \frac{1}{2}\Big)^{n-k} = \binom{n}{k} * \Big(\frac{1}{2}\Big)^{n}$

$$ For $ N = 3, $ there will be 3 coin tosses $ \\\\\\ $ Probability of observing 0 heads $ = \binom{3}{0} * \Big(\frac{1}{2}\Big)^{3} \\\\\\ = \frac{3!}{0! * (3-0)!} * \frac{1}{8} = \frac{1}{8} \\\\\\ $ Probability of observing 1 head $ = \binom{3}{1} * \Big(\frac{1}{2}\Big)^{3} \\\\\\ = \frac{3!}{1! * (3-1)!} * \frac{1}{8} = \frac{3 * 2!}{1 * 2!} * \frac{1}{8} = \frac{3}{8}\\\\\\ $ Probability of observing 2 heads $ = \binom{3}{2} * \Big(\frac{1}{2}\Big)^{3} \\\\\\ = \frac{3!}{2! * (3-2)!} * \frac{1}{8} = \frac{3 * 2!}{2! * 1} * \frac{1}{8} = \frac{3}{8}\\\\ \\ $ Probability of observing 3 heads $ = \binom{3}{3} * \Big(\frac{1}{2}\Big)^{3} \\\\\\ = \frac{3!}{3! * (3-3)!} * \frac{1}{8} = \frac{1}{8}$

$\large $ So, conditional probability mass function of $ X $ given $ N = 3 \\\\ P_{X|N}(x\ |\ 3) = \left\{\begin{matrix} \frac{1}{8}\ ; \ x = 0\\ \\ \frac{3}{8}\ ; \ x = 1\\ \\ \frac{3}{8}\ ; \ x = 2\\ \\ \frac{1}{8}\ ; \ x = 3 \end{matrix}\right. \\\\ --------------------------$

$b) \\\\ $ The number of heads can be $ 5 $ when the coin is tossed at least 5 times $ \\\\ $ Coin can be tossed at least 5 times when $ N = 5 $ or $ N = 6 \\\\ $ When $ N = 5 $ , coin is tossed 5 times and then, \\ \\the probability of obtaing 5 heads $ : P(X = 5 \ | \ N = 5) \\ \\= \binom{5}{5}\Big(\frac{1}{2}\Big)^{5} = \frac{5!}{5!(5-5)!} * \frac{1}{32} = \frac{1}{32} \\\\\\ $ When $ N = 6 $ , coin is tossed 6 times and then, \\ \\the probability of obtaing 5 heads $ : P(X = 5 \ | \ N = 6) \\ \\= \binom{6}{5}\Big(\frac{1}{2}\Big)^{6} = \frac{6!}{5!(6-5)!} * \frac{1}{64} = \frac{6 * 5!}{5! * 1} * \frac{1}{64} = \frac{3}{32} \\\\\\$

$\large P(X = 5) = P(X = 5 $ and $ N=5) + P(X = 5 $ and $ N = 6) \\\\ = P(X = 5\ | \ N = 5) * P(N = 5) + P(X = 5 \ | \ N = 6) * P(N = 6) \\\\ = \frac{1}{32} * \frac{1}{6} + \frac{3}{32} * \frac{1}{6} = \mathbf{\frac{1}{48}} \\\\ -----------------------------$

$\large c) \\\\ $ For y tosses of a fair coin, expected number of heads $ :\\\\ E[X \ |\ N = y] = y * \frac{1}{2} = \frac{y}{2}\\ \\\\ E[X] = \sum_{n=1}^{n=6}\ E[X | N = n] * P(N = n) \\\\ = \frac{1}{6} * \sum_{n=1}^{n=6}\ E[X | N = n] \\\\ = \frac{1}{6} * \sum_{n=1}^{n=6}\ \frac{n}{2} \\\\ = \frac{1}{12} * \sum_{n=1}^{n=6}\ n \\\\ = \frac{1}{12} * \Big( \ \frac{n(n+1)}{2}\ \Big)\Big|_{n=6 } \\\\ = \frac{1}{12} * \frac{6 * (6+1)}{2} = \mathbf{1.75}$