Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws?
P(at least 3 throws) = P(exactly 3 throws) + P(all 4 throws).himu wrote:Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws?
Let W = win and L = lose.
Since P(W) = 1/5, P(L) = 4/5.
Case 1: Leila wins on exactly 3 throws
One way to win on exactly 3 throws is to lose only on the first throw:
P(LWWW) = 1/5 * 1/5 * 1/5 * 4/5 = 4/5�.
This result represents ONE WAY to win on exactly 3 throws.
Now we must account for ALL OF THE WAYS to win on exactly 3 throws.
Since L could be the 1st, 2nd, 3rd, or 4th throw -- for a total of FOUR WAYS -- we multiply by 4:
4 * 4/5� = 16/5�.
Case 2: Leila wins on all 4 throws
P(WWWW) = 1/5 * 1/5 * 1/5 * 1/5 = 1/5�.
Since either Case 1 OR Case 2 will yield a favorable outcome, we ADD the probabilities:
16/5� + 1/5� = 17/5�.
















