Question

A group of 8 children go on a field trip. Mrs. Smith is the mother of 3 of these children. School officials later announce that 2 of the children in the group were exposed to hepatitis while on the trip, but they provide no information as to which 2 children were exposed.
(a) What is the probability that both of the exposed children were Smith children?

(b) What is the probability that exactly one of the two exposed children was a Smith?

(c) What is the probability that at least one of the two exposed children was a Smith?

(d) What is the probability that neither exposed child was a Smith?

(e) One of Mrs. Smith's children in the group is named Amy. What is the probability that Amy Smith was one of the two exposed children?

(f) School officials contact Mrs. Smith to let her know that Amy was one of the two exposed children. What then is the conditional probability that the other exposed child is one of her other children?

Answer 1

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

Step-by-step 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

Answer 2

Answer:

a. 0.107

b. 0.535

c. 0.642

d. 0.4

e. 0.25

f. 0.063

Step-by-step explanation:

Please see attachment