Answer:
a. 3/28 or 0.107
b. 15/28 or 0.536
c. 9/14 or 0.643
d. 5/14 or 0.357
e. 1/4 or 0.25
f. 2/7 or 0.286
Explanation:
a. If three children out of the eight children that went on the field trip are Mrs Smith's then the probability of selecting Mrs Smith's child =
Number of Mrs Smith's children/ total number of children
= 3/8.
Now, if two out of the eight children that went for the field trip were exposed to hepatitis, the probability that both children were Mrs Smith's (without replacement) is:
P (both are Smith)
= 3/8 × 2/7
= 6/56
= 3/28
= 0.107
b. The probability that exactly one out of the two exposed children was Mrs Smith's child.
In this case, one must be a Smith and the other child must not be a Smith. Since three out of the eight children in the group are Mrs Smith's children, then the remaining five children are not Mrs Smith's children.
The probability of selecting a child that is not a Smith is:
Number of children that are not Mrs Smith's/Total number of children
P(exactly one is Smith)
= P(first is Smith,second is not) + p(first is not, second is Smith)
= (3/8 × 5/7) + (5/8 × 3/7)
= 15/56 + 15/56
= 30/56
= 15/28
= 0.536
c. The probability that at least one of the two exposed children was a Smith:
P(at least one is Smith) = 1 - P(the two are not Smith's)
P(the two are not Smith's)
= 5/8 × 4/7
= 20/56
= 10/28
= 5/14
= 0.357
P(at least one is Smith)
= 1 - (5/14)
= 9/14
= 0.643
d. The probability that neither exposed child was a Smith
P(the two are not Smith's)
= 5/8 × 4/7
= 20/56
= 10/28
= 5/14
= 0.357
e. If the name of one of Smith's children who was one of the eight children in the group is Amy, then the probability of selecting Amy should be:
Amy/total number of students.
Amy is just one person therein, the probability of selecting Amy is 1/8.
Now, the probability of selecting any other child that isn't Amy = 7/8
The Probability that Amy was one of the two exposed children is:
P(Amy exposed) = P(Amy,other) + P(other,Amy)
(1/8 × 7/7) + (7/8 × 1/7)
= 1/8 + 1/8
= 2/8
= 1/4
f. It is confirmed that Amy is exposed. We are to find the conditional probability that the other exposed child is also a Smith.
P[another Smith exposed | Amy exposed] = P(Amy & another Smith exposed)/ P(Amy exposed)
P[Amy and another Smith exposed]
= (1C1 × 2C1)/8C2
1C1 = 1!/(0!1!) = 1
2C1 = 2!/(1!1!) = 2
8C2 = 8!/(6!2!) = (8 × 7)/2! = 28
(1C1 × 2C1)/8C2 =
(1 × 2)/28
= 2/28
= 1/14
Remember that P(Amy exposed) = 1/4
P(Amy & another Smith exposed)/P(Amy exposed)
= 1/14 ÷ 1/4
= 1/14 × 4/1
= 4/14
= 2/7
= 0.286
Therefore, P[another Smith exposed | Amy exposed] = 2/7 or 0.286