Question

In a carnival game, there are six identical boxes, one of which contains a prize. Is it appropriate to use the binomial probability distribution to find the probability that a contestant who plays the game five times wins exactly twice?

A contestant wins the prize by selecting the box containing it. Before each game, the old prize is removed and another prize is placed at random in one of the six boxes.

Answer 1

Totally appropriate. `C_(5,2)(1/6)^2(5/6)^(3)~=.1608=16.08%`

Explanation:

The Binomial Probability expression is:

`sum_(k=0)^(n)C_(n,k)(p)^k((~p)^(n-k))`

where `n` is the number of attempts, `k` is successes, `p` is the probability of each success and `~p` is the probability of each failure. We can sum all the attempts to achieve a sum of 1 - showing that we have captured (or can capture) the entirety of the possibilities.

For this question, we're looking specifically at one condition - playing the game 5 times and winning twice. `p=1/6 and ~p=5/6`:

`C_(5,2)(1/6)^2(5/6)^(3)~=.1608=16.08%`