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:
122 comments Page 6 of 13.
Sumanta said:
1 decade ago
See, 3 coins. 3 bits. 3 bits can make numbers from 0-7 is binary:
000, 001, 010, 011, 100, 101, 110, 111.
Take 0->head, 1->tail.
So, max 2 head mean. Not more than 2 heads. We can take samples with 0 head, 1 head and 2 heads.
All except 111.
Thats it.
Total = 8.
Sample = 7.
P = 7/8.
000, 001, 010, 011, 100, 101, 110, 111.
Take 0->head, 1->tail.
So, max 2 head mean. Not more than 2 heads. We can take samples with 0 head, 1 head and 2 heads.
All except 111.
Thats it.
Total = 8.
Sample = 7.
P = 7/8.
Dhruvil said:
1 decade ago
One unbiased coin contains two sides.
So three unbiased coin contain total six sides. From the three coins when tossed together we get {H1H2T3, H1T2H3, T1H2H3}.
So probability is (3/6)= (1/2).
So three unbiased coin contain total six sides. From the three coins when tossed together we get {H1H2T3, H1T2H3, T1H2H3}.
So probability is (3/6)= (1/2).
Lettisha.L.S said:
1 decade ago
Okay, people have been trying to explain it to other people, but.
Why do we consider it as 2^3? I mean the total no. of outcomes, won't it be 9? (3x3) (0.0).
Isn't that how the sample is calculated?
Why do we consider it as 2^3? I mean the total no. of outcomes, won't it be 9? (3x3) (0.0).
Isn't that how the sample is calculated?
Pravin patel said:
1 decade ago
@LETTISHA.
Possibility is head or tail, mean 2 time.
And it is to 3 time means 2^3 = 8(2*2*2) = 8.
Hear is coin, if we are using dies then 6^3.
Possibility is head or tail, mean 2 time.
And it is to 3 time means 2^3 = 8(2*2*2) = 8.
Hear is coin, if we are using dies then 6^3.
Manan said:
1 decade ago
Can't it be like this:
Probability of all events = 3 heads, 2 heads, 1 head, 0 heads.
Hence n(s) = 4.
n(e) = 3.
Probability = 3/4.
Probability of all events = 3 heads, 2 heads, 1 head, 0 heads.
Hence n(s) = 4.
n(e) = 3.
Probability = 3/4.
Sonali malviya said:
1 decade ago
Can anybody tell me why we have taken n(e) = 7?
Ghanshyam said:
1 decade ago
How P(E) = n(E) upon n(S)= 7 by 8.
Riems said:
1 decade ago
Is there any direct method for calculating it? Because calculating n(e) may get confused in the exams time.
Brian said:
1 decade ago
What is the probability of this question?
Joel said:
1 decade ago
At most 2 heads, so how did you include HHH?
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