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John flips a coin 4 times. What is the probability

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by M7MBA » Sat Dec 23, 2017 3:37 am
John flips a coin 4 times. What is the probability that he gets heads on at least one of the four flips?

(A) 1/16
(B) 1/4
(C) 1/2
(D) 3/4
(E) 15/16

The OA is E.

I don't know how to solve this PS question. Experts, may you give me an explanation? I'd be thankful.
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by elias.latour.apex » Mon Dec 25, 2017 1:45 pm
Well, one way to solve this problem is to figure his chance of tossing one heads, his chance of tossing two heads, his chance of tossing three heads, and his chance of tossing four heads and adding all those numbers up.

An easier way is to figure his chance of tossing all tails. If he doesn't toss all tails, then he must have tossed at least one heads.

The chance of tails four times is (1/2)^4 or 1/16 that he will toss nothing but tails.

Accordingly, he must have a 15/16 chance of tossing heads at least once.
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by Scott@TargetTestPrep » Thu Sep 05, 2019 5:17 am
M7MBA wrote:John flips a coin 4 times. What is the probability that he gets heads on at least one of the four flips?

(A) 1/16
(B) 1/4
(C) 1/2
(D) 3/4
(E) 15/16

The OA is E.

I don't know how to solve this PS question. Experts, may you give me an explanation? I'd be thankful.
We can use the formula:

P(at least one head on 4 flips) = 1 - P(no heads on 4 flips)

P(no heads on 4 flips) = (1/2)^4 = 1/16.

P(at least one head on 4 flips) = 1 - 1/16 = 15/16.

Answer: E

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by Scott@TargetTestPrep » Thu Sep 05, 2019 5:17 am
Scott@TargetTestPrep wrote:
M7MBA wrote:John flips a coin 4 times. What is the probability that he gets heads on at least one of the four flips?

(A) 1/16
(B) 1/4
(C) 1/2
(D) 3/4
(E) 15/16
We can create the equation:

(a^2) x (-3) = 3(3a + 2)

-a^2 = 3a + 2

-a^2 - 3a - 2 = 0

a^2 + 3a + 2 = 0

(a + 2)(a + 1) = 0

a = -2 or a = -1

Answer: B

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by Brent@GMATPrepNow » Thu Sep 05, 2019 6:01 am
Hey Scott,

I think you posted your answer on the wrong thread.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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