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

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Joined: 04 Mar 2012
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A fair coin is tossed 4 times. What is the probability of  [#permalink]

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30 Apr 2012, 01:05
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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
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Posts: 47961
Re: A fair coin is tossed 4 times. What is the probability of  [#permalink]

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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).

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

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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01 Sep 2013, 19:41
2
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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01 Sep 2013, 23:31
1
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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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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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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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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A fair coin is tossed 4 times. What is the probability of  [#permalink]

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

The concept is binomial probabilities.

Proba of having k success out-of n events for a binomial variable has a given 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]

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04 Dec 2017, 04:12
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

I eliminated the options like follows:

A coin is tossed 4 times,
Probability of getting "at least 2" can be getting 2 tails, 3 tails and 4 tails.

Thus, probability >0.5.

There is only one option with probability >0.5

Please correct me if I am wrong, and yes I agree this is very crude approach.

I am very weak in probability.
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Re: A fair coin is tossed 4 times. What is the probability of  [#permalink]

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12 Jan 2018, 07:31
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

P(at least 2 tails) = 1 - P(not getting at least 2 tails)

P(at least 2 tails) = 1 - [P(getting exactly 1 tail) + P(getting 0 tails)]

Let’s look at P(getting exactly 1 tail):

P(T - H - H - H) = (1/2)^4 = 1/16

Since T - H - H - H can be arranged in 4!/3! = 4 ways, then P(getting exactly 1 tail) = 4 x 1/16 = 4/16.

Now let’s look at P(getting 0 tails), i.e., P(getting all 4 heads):

P(H - H - H - H) = (1/2)^4 = 1/16

P(at least 2 tails) = 1 - (4/16 + 1/16) = 11/16.

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

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12 Jan 2018, 15:22
Total outcome = 2^4 = 16

3 cases for favorable outcomes: (HHTT), (HTTT), (TTTT)
4C2 + 4C3 + 4C4 => 6 + 4 + 1= 11

Probability of at least 2 = (favorable outcome/ total outcome) = 11/16
A fair coin is tossed 4 times. What is the probability of &nbs [#permalink] 12 Jan 2018, 15:22
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