Archit143 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?
A. 1/5^4
B. 1/5^3
C. 6/5^4
D. 13/5^4
E. 17/5^4
Official Solution (Credit: Manhattan Prep)
There are two ways that Leila can succeed on at least 3 of the throws—she can succeed on exactly 3 throws or on all 4. Find the total probability by determining the chance of each of these two distinct outcomes and adding them together.
Start with the chance of succeeding on all 4 throws. Since Leila’s chance of success is 1/5, multiply out her chance of succeeding 4 times in a row:
1/5 × 1/5 × 1/5 × 1/5 = 1/54
You can now eliminate answer A; it can’t be correct, because you haven’t yet added in the possibility of succeeding on exactly 3 throws.
If Leila succeeds on exactly 3 of the throws, then she must have 1 missed throw. The chance of missing a throw is 1 – 1/5 = 4/5, so multiply the individual probabilities to get the total probability:
1/5 × 1/5 × 1/5 × 4/5 = 4/54
However, this is the probability of only one specific outcome: hit, hit, hit, miss. Since she doesn’t need to make her throws in this specific order, you'll also need to consider the other possibilities:
hit, hit, miss, hit
hit, miss, hit, hit
miss, hit, hit, hit
There are 4 different ways that Leila can get 3 hits and 1 miss. (Note that this wasn’t an issue when calculating the chance of succeeding on all 4 throws, because there is only one way to do that.) Therefore, her chance of successfully making exactly 3 out of 4 throws is 4(4/54)=16/54.
Finally, add the two values together to find the probability that Leila succeeds on at least 3 of the throws:
1/15^4+16/15^4 = 17/15^4
The correct answer is E.