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Probability (using complement)

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by NDiwan » Sat Aug 08, 2015 10:23 pm
Experts,

Could you help me figure out this problem? Can it be solved using the complement formula?

-- 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
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by Uva@90 » Sat Aug 08, 2015 11:19 pm
NDiwan wrote:Experts,

Could you help me figure out this problem? Can it be solved using the complement formula?

-- 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
Hi NDiwan,

Prob of success = 1/5
prob of Failure = 4/5

No of chances = 4

To find: At least 3 of the throws

So it can 3 out of 4 OR 4 out of 4 throws

1) 3 out of 4

SSSF or SSFS or SFSS or FSSS

SSSF = (1/5) *(1/5) *(1/5) * (4/5) = 4/5^4
there are 4 ways we can arrange them, so 4*4/5^4 = 16/5^4

2) 4 out of 4 throws
SSSS =
(1/5) *(1/5) *(1/5) * (1/5) = 1/5^4

add 1 and 2

17/5^4

Answer is E

Regards,
Uva.
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by GMATGuruNY » Sun Aug 09, 2015 2:40 am
NDiwan wrote:A) 1/5^4

B) 1/5^3

C) 6/5^4

D) 13/5^4

E) 17/5^4
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�.

The correct answer is E.
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by Brent@GMATPrepNow » Sun Aug 09, 2015 6:50 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
Last edited by Brent@GMATPrepNow on Mon Aug 29, 2016 6:40 am, edited 1 time in total.
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by NDiwan » Sun Aug 09, 2015 10:17 am
Thank you, all! Appreciate your responses.

I have to mention, probability is not my strongest area, but thanks to Brent's PrepNow videos, it seems like I'm starting to gather more than I have ever understood about this topic. So, thanks Brent! :)
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