Probability
Aptitude

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Q.

When 3 fair coins are tossed together, what is the probability of getting atleast 2 tails?

 A.

1/4

 B.

2/3

 C.

1/3

 D.

1/2

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Solution:
Option(D) is correct

We can get 2 tails from 3 trials in \(^3C_2\) ways. 

We can get 3 tails in 3 trials in exactly one way.

Thus, required probability 

\(=\dfrac{(3+1)}{8}\)

\(=\dfrac{1}{2}\)

Edit: Thank you Shruthi Bhaasini for pointing put the anomoly, modified the question for 'atleast 2 tails'.


(9) Comment(s)


Mike
 ()

As you are flipping the coin only 3 times, the case where their is an equal number of heads and tails can be discussed. Thus, there will ALWAYS be the case that you have at least 2 heads or 2 tails. As the coins are fair, it is an equal chance that it will be heads or tails. Thus the answer is simply 1/2.



Shruthi Bhaasini
 ()

The question should be the probability of getting atleast two tails.


Deepak
 ()

Thank you Shruthi for pointing out, modified the question.


Nayyer
 ()

The answer could be $\dfrac{3}{8}.

Question wording is not clear.



Piyush
 ()

Excatly what ravish say about the question



Dhanna
 ()

Possible outcomes are:

1: TTT
2: TTH
3: THT
4: HTT
5: THH
6: HTH
7: HHT
8: HHH

out of the possible 8 outcomes only 4 (TTT, TTH, THT, HTT) are favourable.

So required probability $= \dfrac{4}{8}=\dfrac{1}{2}$



Ravish
 ()

it is not given as at least 2 tails in the question.

The answer should be $\dfrac{3}{8}$ as per the question.


Dhanna
 ()

Ravish it does not say EXACTLY 2 tails. If we get all the three tails as an outcome then also it is a valid outcome.

So I guess question is correct.