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 7 of 13.
Shubham said:
9 years ago
Here they've taken 3 coins for example. Why not 2 coins?
Megha said:
9 years ago
Thank you very much @Sundar.
Sambit said:
9 years ago
But in explaining in the event there is TTT is included that is the wrong HHH is included is correct.
Raj@RS@ said:
9 years ago
What is the probability, that a number selected from 1, 2, 3, --- 2, 5, is a prime number, when each of the numbers is equally likely to be selected.
Can anyone solve this?
Can anyone solve this?
Trupti said:
9 years ago
Why zero will be accepted? Please tell me.
Chinenye said:
9 years ago
Why count 0, 0 means nothing?
Jyotsna said:
9 years ago
Thank you @Sundar.
Your explanation is clear and simple.
Your explanation is clear and simple.
XYZ said:
9 years ago
Can anyone tell me the combination of three unbiased coins?
Rushikesh said:
9 years ago
How should we accept as zero heads?
Sarma said:
10 years ago
3/4 is the answer.
Why are you considering TTT as a possible case?
Why are you considering TTT as a possible case?
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