Question

six faces fair dice are consider 1,2,2,3,3,3 and an unbiased coin is thrown the number of times indicated by the score on the dice . let H denote number of heads obtained, prove that P(H= 2) = 13/48

Answer 1

I'm assuming this experiment plays out by first rolling the die, and whatever face value occurs determines the number of coin tosses. Let
`X` denote the random variable for the face value of the die. Then the PMF for
`X` is


`\mathbb P(X=x)=\begin{cases}\frac16&\text{for }x=1\\\\\frac13&\text{for }x=2\\\\\frac12&\text{for }x=3\\\\0&\text{otherwise}\end{cases}`

If
`X=1`, then only one coin is flipped, so
`\mathbb P(H=2|X=1)=0` because getting two of any side of the coin is impossible.

If
`X=2`, then the probability of getting two heads is
`\mathbb P(H=2|X=2)=\dbinom22\left(\frac12\right)^2\left(\frac12\right)^0=\frac14`.

If
`X=3`, then the probability of getting two heads is
`\mathbb P(H=2|X=3)=\dbinom32\left(\frac12\right)^2\left(\frac12\right)^1=\frac38`.

By the law of total probability,


`\mathbb P(H=2)=\displaystyle\sum_(x=1)^3\mathbb P(H=2|X=x)\mathbb P(X=x)`

`\mathbb P(H=2)=0*\frac16+\frac14*\frac13+\frac38*\frac12`

`\mathbb P(H=2)=\frac1{12}+\frac3{16}=(13)/(48)`