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Leila's carnival game

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by himu » Mon Feb 11, 2013 8:01 pm
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?
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by GMATGuruNY » Mon Feb 11, 2013 8:26 pm
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?
P(at least 3 throws) = P(exactly 3 throws) + P(all 4 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�.
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by Scott@TargetTestPrep » Fri Dec 15, 2017 7:06 am
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.

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by Brent@GMATPrepNow » Fri Dec 15, 2017 7:21 am
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�

B)1/5³

C)6/5�

D)13/5�

E)17/5�

OAE
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
= 1/5�

P(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
= 4/5�
Keep 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 4/5³
So, P(succeeds 3 times) = 4/5� + 4/5� + 4/5� + 4/5�
= 16/5�

So, P(succeeds AT LEAST 3 times) = P(succeeds 4 times) + P(succeeds 3 times)
= 1/5� + 16/5�
= [spoiler]17/5�[/spoiler]
= E


Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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