SOLUTION: Hi,
Question:
Mrs. Marx has four children. If two of the children are males, what is the probability that at least one of the remaining children is not a male.
Answer (suppos
Algebra ->
Probability-and-statistics
-> SOLUTION: Hi,
Question:
Mrs. Marx has four children. If two of the children are males, what is the probability that at least one of the remaining children is not a male.
Answer (suppos
Log On
Question 1016589: Hi,
Question:
Mrs. Marx has four children. If two of the children are males, what is the probability that at least one of the remaining children is not a male.
Answer (supposedly): 10/11
You can put this solution on YOUR website!
Question:
Mrs. Marx has four children. If two of the children are males, what is the probability that at least one of the remaining children is not a male.
Answer (supposedly): 10/11
Solution:
First, we must realize that this is a case of conditional probability.
There are four children involved, so the sample space is 16 (=2^4).
Let
N = event of having N male children out of the four.
Then by building a probability tree, or by a contingency table, we have
P(0)=1/16
P(1)=4/16
P(2)=6/16
P(3)=4/16
P(4)=1/16
Now return to the required probability:
Probability of having at least one female given at least 2 out of 4 children are male, which is a case of conditional probability.
Conditional probability
P(A|B)=P(A∩B)/P(B)
where P(A|B) reads
"probability of event A given that event B has already happened."
We further define
A = event of having at least one female child (desired event).
B = event of having at least two male children (given condition)
But
P(A∩B)=P(N<4 ∧ N>=2)=(6+4)/16=10/16, and
P(B)=P(N>=2)=(6+4+1)/16=11/16
Therefore
P(A∩B | B)
=P(A∩B)/P(B)
=
=
