Aptitude Discussion

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 |

**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'.

**Mike**

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**Shruthi Bhaasini**

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

Thank you Shruthi for pointing out, modified the question.

**Nayyer**

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

Question wording is not clear.

**Piyush**

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Excatly what ravish say about the question

**Dhanna**

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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**

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

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

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.

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.