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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]

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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
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Re: A fair coin is tossed 4 times. What is the probability of [#permalink]

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New post 30 Apr 2012, 01:17
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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).

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

Answer: D.

Hope it's clear.
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Re: A fair coin is tossed 4 times. What is the probability of [#permalink]

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Re: A fair coin is tossed 4 times. What is the probability of [#permalink]

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New post 01 Sep 2013, 19:41
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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]

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New post 01 Sep 2013, 23:31
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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


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]

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Re: A fair coin is tossed 4 times. What is the probability of [#permalink]

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New post 14 Aug 2015, 00:39
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


Total outcome is 2^4=16
Number of cases of( tails < or =2) = 5, HHHH, THHH, HTHH, HHTH, HHHT
Therefore, the required probability is 11/16
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Re: A fair coin is tossed 4 times. What is the probability of [#permalink]

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New post 08 Dec 2016, 00:47
Probability of 2T and 2H = 4C2*1/16 = 6/16
Probability of 3T and 1H = 4C3*1/16 = 4/16
Probability of 4T and 0H = 4C4*1/16 = 1/16

Summing the probabilities for each scenario that fulfils our requirements = 11/16
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Re: A fair coin is tossed 4 times. What is the probability of [#permalink]

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New post 16 Apr 2017, 00:40
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



This is how I solved it, please tell me if it is correct.

Let's list the possibilities:
HHTT
HTHT
THHT
TTHH
THTH
HTTH

TTTH
TTHT
THTT
HTTT

TTTT

Now let's look at the first one: HHTT. Probability of getting H=1/2, H=1/2, T=1/2, T=1/2. So probability of getting HHTT = 1/2*1/2*1/2*1/2 = 1/16.
Similarly, for all the rest, the probability is 1/16. There are 11 possibilities on the list so the answer is 11*1/16 = 11/16.
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Re: A fair coin is tossed 4 times. What is the probability of [#permalink]

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New post 22 May 2017, 09:10
This is a standard question, which should be approach with a general concept.
Otherwise, it is a waste of time to reconstruct the logic everytime we face this type of question.

The oncept is binomial variable probabilities.

Proba of having k success out-of n events for a binomial variable is given by a formula.
Let say, proba of success = p and proba of failure = (1-p)

Formula is nCk * (p^k)*(1-p)^(n-k)
if p=1/2 then formula becomes nCk/(2^n)

In the current question, p=1/2 and (1-p)=1/2
n=4 and p=2 or 3 or 4

Answer is (4C2+ 4C3 + 4C4)/(2^4) = 11/16

Option - D
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Re: A fair coin is tossed 4 times. What is the probability of   [#permalink] 22 May 2017, 09:10
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