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}.
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n(E) | = | 7 | . |
n(S) | 8 |
Discussion:
120 comments Page 3 of 12.
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?
Awaneesh Kumar said:
6 years ago
I do not understand about this answer, because most two heads coming from events.
I think the ans is 3/8.
TTT in not come head
TTH
THT
HTT
have not come head most times.
I think the ans is 3/8.
TTT in not come head
TTH
THT
HTT
have not come head most times.
HARINI said:
8 years ago
s = {HHH,HHT,HTH,HTT,THT,TTH,TTT}
S(N) = 8.
LET A = GETTING AT MOST 2 HEADS.
(MAX 2 HEADS;MIN 1HEADS OR 0 HEADS)
A ={HHT,HTT,THH,HTT,THT,TTH,TTT}
N(A) = 7,
P(A) = 7\8.
S(N) = 8.
LET A = GETTING AT MOST 2 HEADS.
(MAX 2 HEADS;MIN 1HEADS OR 0 HEADS)
A ={HHT,HTT,THH,HTT,THT,TTH,TTT}
N(A) = 7,
P(A) = 7\8.
Sujit said:
1 decade ago
SOURAV
If NOW will be together all the time all the time then take "NOW" as one letter. Lucknow will be considered as 5 letters. Then you can get the answer easily.
If NOW will be together all the time all the time then take "NOW" as one letter. Lucknow will be considered as 5 letters. Then you can get the answer easily.
Amit Verma said:
2 years 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.
(37)
Reena said:
10 months ago
At most 2 heads means it cannot be more than 2. It can be zero or 1 or 2 but not 3. And there is only one case that have 3 (HHH). So probability is 7/8.
(26)
Apti's don said:
1 decade ago
Simple and smart solution:
Find probability of getting 3 head. then subtract it from 1.
2C1*2C1*2C1 = 8.
Now 1/8 is probability,
So 1 - 1/8 = 7/8.
Find probability of getting 3 head. then subtract it from 1.
2C1*2C1*2C1 = 8.
Now 1/8 is probability,
So 1 - 1/8 = 7/8.
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.
Atul Kushwaha said:
1 decade ago
If the question would be to find out the the probability of getting at least one head then it must be 7/8 in this case also. Thanks :"Indiabix".
Yadlapalli sivateja said:
9 years ago
To HHH condition:
H (1/2).
H (1/2)+3!/3! = 1/8.
H (1/2) answer = 1/8+3/8 = 1/2.
To TTH condition:
T(1/2).
T(1/2)+3!/2! = 3/8.
T (1/2).
H (1/2).
H (1/2)+3!/3! = 1/8.
H (1/2) answer = 1/8+3/8 = 1/2.
To TTH condition:
T(1/2).
T(1/2)+3!/2! = 3/8.
T (1/2).
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