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?
3
4
1
4
3
8
7
8
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:
122 comments Page 13 of 13.

Sriram said:   6 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.

Dhivya said:   5 years ago
Thank you for explaining @Sundar.


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