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# A fair coin is tossed 4 times. What is the probability of

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A fair coin is tossed 4 times. What is the probability of [#permalink]  30 Apr 2012, 00:05
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A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8
[Reveal] Spoiler: OA
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Re: A fair coin is tossed 4 times. What is the probability of [#permalink]  30 Apr 2012, 00:17
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gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

Let's find the probability of the opposite event and subtract this value from 1.

The opposite event would be getting zero tails (so all heads) or 1 tail.

$$P(HHHH)=(\frac{1}{2})^4=\frac{1}{16}$$.

$$P(THHH)=\frac{4!}{3!}*(\frac{1}{2})^4=\frac{4}{16}$$, we are multiplying by $$\frac{4!}{3!}$$ since THHH scenario can occur in number of ways: THHH, HTHH , HHTH, or HHHT (notice that $$\frac{4!}{3!}$$ basically gives number of arrangements of 4 letters THHH out of which 3 H's are identcal).

$$P(T\geq{2})=1-(\frac{1}{16}+\frac{4}{16})=\frac{11}{16}$$.

Hope it's clear.
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Re: A fair coin is tossed 4 times. What is the probability of [#permalink]  13 Jul 2013, 23:13
Expert's post
Bumping for review and further discussion.
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Re: A fair coin is tossed 4 times. What is the probability of [#permalink]  01 Sep 2013, 18:41
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gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

Another method:

Let's find the probability of 2 tails , 3 tails and all 4 tails

P(TTTT)=((1/2)^4=1/16.

P(HTTT)=(4!/3!)*(1/2)^4=4/16, we are multiplying by 4C3 (notice that {4!/3!} basically gives number of arrangements of 4 letters HTTT out of which 3 T's are identcal).

P(TTHH) = 4C2*(1/2)^4 = 4!/2!*2!*(1/2)^4=6/16

Total Probablity = 1/16 + 4/16 + 6/16
=11/16
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Re: A fair coin is tossed 4 times. What is the probability of [#permalink]  01 Sep 2013, 22:31
gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

Here's another approach that I used:

Total possibilities = 2^4= 16 since each toss leads to 2 outcomes.

Way of getting no tail in 4 tosses = 1 (H H H H)
Way of getting 1 tail in 4 tosses = 4 (T H H H) (H T H H) ( H H T H) (H H H T)
So ways of getting at least two tails in 4 tosses = 11 (16-5)
Therefore, the probability = 11/16 (D).
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Re: A fair coin is tossed 4 times. What is the probability of [#permalink]  23 Mar 2015, 10:47
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A fair coin is tossed 4 times. What is the probability of   [#permalink] 23 Mar 2015, 10:47
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