A fair coin and fair die
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)
Welcome to Math Help Forum!
The coin can land in ways each time it is tossed. So there are ways is which the coin can land if it is tossed times.
Originally Posted by sobad
The die can land in ways, and so the number of elements in the equiprobable sample space is , each element consisting of a sequence of H's and T's followed by a score from to .
The distribution of the number of heads is Binomial, and the probability of getting heads (where ) is:
The probability that the score on the die, , is less than is given by:
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 .
According to my calculations (using a spreadsheet to do the arithmetic) this comes out as to 4 d.p.