A fair coin and fair die

Hello sobad

Welcome to Math Help Forum!

Quote:

Originally Posted by sobad

Any help would be appreciated!

A fair coin is tossed 6 times--then 1 fair die is rolled. Set up an equiprobable sample space & calculate the prob. that the number on the die is less than the number of heads tossed

No arithmetic is necessary because the answer would be crazy long.

Thank you in advance (Nod)

The coin can land in $2$ ways each time it is tossed. So there are $2^6=64$ ways is which the coin can land if it is tossed $6$ times.

The die can land in $6$ ways, and so the number of elements in the equiprobable sample space is $64\times 6 = 384$ , each element consisting of a sequence of H's and T's followed by a score from $1$ to $6$ .

The distribution of the number of heads is Binomial, and the probability of getting $r$ heads (where $0\le r\le 6$ ) is:

$p(r) = \binom{6}{r}\Big(\frac12\Big)^6$

The probability that the score on the die, $s$ , is less than $r$ is given by:

$p(s<0)=0$
$p(s<1)=0$
$p(s<2)=\tfrac16$
$p(s<3)=\tfrac26$
$p(s<4)=\tfrac36$
$p(s<5)=\tfrac46$
$p(s<6)=\tfrac56$

Or: $p(s<r) = \left\{ <br /> \begin{array}{l l}<br /> 0 & r\le 1\\<br /> \dfrac{r-1}{6} & 2\le r\le 6\\<br /> \end{array} \right.$

So to calculate the probability that, for any given throw of the coins, the score on the die is less than the number of heads, you'll need to multiply the two respective probabilities together, and then add all the resulting products.

As a summation formula, this is $\sum_{r=0}^6\binom{6}{r}\Big(\frac12\Big)^6p(s<r)$ .

According to my calculations (using a spreadsheet to do the arithmetic) this comes out as $0.3359$ to 4 d.p.

Grandad