GeminiHeat 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. \(\frac{1}{5^4}\)
B. \(\frac{1}{5^3}\)
C. \(\frac{6}{5^4}\)
D. \(\frac{13}{5^4}\)
E. \(\frac{17}{5^4}\)
Given: P(succeeds on 1 throw) = 1/5
P(succeeds at least 3 times) = P(succeeds 4 times
OR succeeds 3 times)
= P(succeeds 4 times)
+ P(succeeds 3 times)
P(succeeds 4 times)P(succeeds 4 times) = P(succeeds 1st time
AND succeeds 2nd time
AND succeeds 3rd time
AND succeeds 4th time)
= P(succeeds 1st time)
x P(succeeds 2nd time)
x P(succeeds 3rd time)
x P(succeeds 4th time)
= 1/5
x 1/5
x 1/5
x 1/5
=
1/5⁴P(succeeds 3 times)Let's examine
one possible scenario in which Leila succeeds exactly 3 times:
P(FAILS the 1st time
AND succeeds 2nd time
AND succeeds 3rd time
AND succeeds 4th time)
= P(FAILS the 1st time)
x P(succeeds 2nd time)
x P(succeeds 3rd time)
x P(succeeds 4th time)
= 4/5
x 1/5
x 1/5
x 1/5
=
4/5⁴Keep in mind that this is only ONE possible scenario in which Leila succeeds exactly 3 times (Leila fails the 1st time).
Leila can also FAIL the 2nd time, or the 3rd time or the 4th time.
Each of these probabilities will also equal
4/5³So, P(succeeds 3 times) =
4/5⁴ +
4/5⁴ +
4/5⁴ +
4/5⁴=
16/5⁴So, P(succeeds AT LEAST 3 times) = P(succeeds 4 times)
+ P(succeeds 3 times)
=
1/5⁴ +
16/5⁴= 17/5⁴
Answer: E
04 May 2021, 03:36 
61% (01:59) correct
