# Aptitude - Probability - Discussion

Discussion Forum : Probability - General Questions (Q.No. 5)

5.

Three unbiased coins are tossed. What is the probability of getting at most two heads?

Answer: Option

Explanation:

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

P(E) = | n(E) |
= | 7 | . |

n(S) |
8 |

Discussion:

119 comments Page 2 of 12.
Amit Verma said:
1 year ago

Here S = {TTT, TTH, THT, THH, HHH, HHT, HTH, HTT}

Let E = Even of getting at most two heads.

Then E = {HHH, HHT, HTH, THH},

P(E) = n(E)/n(S),

= 4/8 = 1/2.

Let E = Even of getting at most two heads.

Then E = {HHH, HHT, HTH, THH},

P(E) = n(E)/n(S),

= 4/8 = 1/2.

(36)

Tejeshwari said:
2 years ago

Thanks all for explaining it.

(2)

Zahid said:
2 years ago

How to find the events of coins quickly like TTT and HHT etc? explain me.

(5)

MUKESH MAHLI said:
3 years ago

At most mean- less than or less than equal to.

Am I right?

Am I right?

(4)

Dhivya said:
3 years ago

Thank you for explaining @Sundar.

Sumanth said:
4 years ago

It is 6/8=3/4.

(4)

Meena said:
4 years ago

The question is what is the probability of getting "at most" two heads?

Here, at most two heads mean Not more than two heads.

So, HHH is more than two heads. It is not counted in Event. Am I right?

Here, at most two heads mean Not more than two heads.

So, HHH is more than two heads. It is not counted in Event. Am I right?

(6)

Neeraja said:
4 years ago

@ Vivek Rai.

A coin is tossed 3 times. We have 8 possible outcomes. In those 8 possible outcomes, we should choose the outcomes in which we have 1 head, 2 heads or no heads (3tails). But according to you if you consider the only head side of the coin then how did you get 8 in the denominator?

So the answer is 7/8.

A coin is tossed 3 times. We have 8 possible outcomes. In those 8 possible outcomes, we should choose the outcomes in which we have 1 head, 2 heads or no heads (3tails). But according to you if you consider the only head side of the coin then how did you get 8 in the denominator?

So the answer is 7/8.

(5)

Vivek Rai said:
4 years ago

But it says heads that mean only head should be counted not the tail side of the coin as we know a coin has 2 side head and tail and the question is asking about at most 2 heads so it should be = 3/8.

Anyone, please clarify my doubts.

Anyone, please clarify my doubts.

(2)

Sriram said:
4 years ago

@sara.

We can get 2 heads any position out of 3 (- - -) each position has 1/2 probability. So, (1/2*1/2*1/2) + (1/2*1/2*1/2) + (1/2*1/2*1/2) =3/8.

Do the same way for 1 head (1 out of 3 positions) is (1/2*1/2*1/2) + (1/2*1/2*1/2) + (1/2*1/2*1/2) =3/8.

And last without a head is 1/8.

Add (atmost 2 to atlast 0) 1/8+3/8+3/8=7/8.

We can get 2 heads any position out of 3 (- - -) each position has 1/2 probability. So, (1/2*1/2*1/2) + (1/2*1/2*1/2) + (1/2*1/2*1/2) =3/8.

Do the same way for 1 head (1 out of 3 positions) is (1/2*1/2*1/2) + (1/2*1/2*1/2) + (1/2*1/2*1/2) =3/8.

And last without a head is 1/8.

Add (atmost 2 to atlast 0) 1/8+3/8+3/8=7/8.

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