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:
121 comments Page 10 of 13.
Arun Kumar said:
7 years ago
@All.
Note : At most means upto two heads not more than it.
Note : At most means upto two heads not more than it.
Kharkhanhimel said:
6 years ago
Thanks all.
Javeed said:
6 years ago
In coins, we have 2 sides so 2, and how many coins they gave in the question is 3 so 2^3 = 8.
And the number HH is 7 so 7/8 is answer.
And the number HH is 7 so 7/8 is answer.
Madhu said:
6 years ago
What is unbiased coins?
Can anyone please answer me.
Can anyone please answer me.
Prashansa said:
6 years ago
I am not understanding this. Please, someone, help me.
Lingesh kumar said:
6 years ago
What is unbaised coins?
(1)
Aman Gupta said:
6 years ago
At most two head means that 2 or more than 2 heads are accepted but no coins are not haed or one head are not accepted.
Jaya said:
6 years ago
Thank you for your explanation @Sundhar.
Upendra said:
5 years ago
Could we use binomial probability distribution?
Sriram said:
5 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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