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 10 of 13.
Shah Rukh said:
9 years ago
Biased means neither head nor tail.
Unbiased means opposite to biased. i.e It will be in perfect outcomes.
Unbiased means opposite to biased. i.e It will be in perfect outcomes.
HARINI said:
9 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.
Praveen said:
9 years ago
Give the sample space of the given problem.
Praveen kumar said:
9 years ago
You are great, Thanks @Sundar.
Deva said:
9 years ago
Thank You @SUNDAR.
Shubham said:
9 years ago
Thanks @Harini.
Shaikh Sayma said:
9 years ago
What is the Probability of getting at the most 2 tails?
Akhil said:
8 years ago
@Belle.
A. P(All boys) = (1/2)^3 = 1/8.
B. P(All girls or all boys) = 1/8 + 1/8 = 1/4.
C. P(Exactly two boys or two girls) = 2*3C2(1/2)^2*(1/2) = 2*3*(1/8) = 3/4(here we need 2 of particular gender. So., if we calculate[2*3C2(1/2)^2] we'll get 2 of particular gender and then this should be multiplied with 1/2. Atlast we get 3/4).
D. P(At least one child of each gender) = 1 -P(all girls or all boys)= 1 - 1/4 = 3/4.
I hope this helps.
A. P(All boys) = (1/2)^3 = 1/8.
B. P(All girls or all boys) = 1/8 + 1/8 = 1/4.
C. P(Exactly two boys or two girls) = 2*3C2(1/2)^2*(1/2) = 2*3*(1/8) = 3/4(here we need 2 of particular gender. So., if we calculate[2*3C2(1/2)^2] we'll get 2 of particular gender and then this should be multiplied with 1/2. Atlast we get 3/4).
D. P(At least one child of each gender) = 1 -P(all girls or all boys)= 1 - 1/4 = 3/4.
I hope this helps.
Krishna said:
8 years ago
Can someone help me to solve this?
If head appears consecutively in the first three tosses of a fair/unbiased coin. What is a probability of Head appearing in the fourth toss also?
If head appears consecutively in the first three tosses of a fair/unbiased coin. What is a probability of Head appearing in the fourth toss also?
Soaib said:
8 years ago
Thanks @Harini.
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