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?
We need to determine the probability that Leila succeeds on exactly 3 throws or on all 4 throws.
Scenario 1: succeeds on exactly 3 throws
We can let Y denote a successful throw and N denote a non-successful throw:
P(Y-Y-Y-N) = 1/5 x 1/5 x 1/5 x 4/5 = 4/(5^4)
However, we must account for the order of Y-Y-Y-N. Using our formula for indistinguishable items, Y, Y, Y, and N can be arranged in 4!/3! = 4 ways.
Thus, the probability of succeeding on exactly 3 throws (out of 4 attempts) is 4/(5^4) x 4 = 16/(5^4).
Now we can determine scenario 2:
Scenario 2: succeeds on all 4 attempts
P(Y-Y-Y-Y) = 1/5 x 1/5 x 1/5 x 1/5 = 1/(5^4)
Thus, the probability of succeeding on 3 throws out of 4 or 4 throws out of 4 is 16/(5^4) + 1/(5^4) = 17/5^4.
