Math 3C Homework 3 Solutions

Since P (A|B1) = 1, P (A|B2) = 1
2 , P (A|B3) = 0,
P (A) = 1 · 1 1 1 1 1
+ · + 0 · =
3 2 3 3 2 .
21. You are dealt on card from a standard deck of 52 cards. If A denotes the event that the card is a spade
and if B denotes the event that the card is an ace, determine whether A and B are independent.
Solution
P (A) = 1
4
1
1
and P (B) = 13 . The probability P (A ∩ B) that the card is the spade ace is 52 . Hence
and A and B are independent.
P (A ∩ B) = 1
= P (A)P (B)
52
22. You are dealt two cards from a standard deck of 52 cards. If A denotes the event that the first card is an
ace and B denotes the event that the second card is an ace, determine whether A and B are independent.
Solution
From problem 15, we know that P (A) = P (B) = 1
13
aces is (4 2) 4·3 1
= 52·51 = 13·17 . Hence
( 52
2 )
and A and B are not independent.
P (A ∩ B) = P (A)P (B)
. The probability P (A ∩ B) that the two cards are
29. Assume that the probability that an insect species lives more than five days is 0.1. Find the probability
that in a sample of size 10 of the this species at least one insect will still be alive after 5 days.
Solution
The probability that an insect lives more than five days is independent of the other insects. So the
probability that all of the ten insects will die within 5 days is (0.9) 10 . The probability that at least one
insect will be alive after 5 days is therefore 1 − (0.9) 10 .
31. A screening test for a disease shows a positive result in 95% of all cases when the disease is actually present
and in 10% of all cases when it is not. If the prevalence of the disease is 1 in 50, and an individual tests
positive, what is the probability that the individual actually has the disease?
Solution
Let A denote the probability that the individual has the disease and B denote the probability that the
test shows a positive result. Then we need to find P (A|B) =
Therefore, P (A|B) = 0.019
0.117 ≈ 0.1624.
P (A ∩ B) = 0.95 · 1
= 0.019,
50
P (B) = P (B|A)P (A) + P (B|A C )P (A C )
= 0.95 · 1
+ 0.1 ·
50
1 − 1
50
P (A∩B)
P (B) . Since the prevalence is 1
50 ,
= 0.117.
32. A screening test for a disease shows a positive result in 95% of all cases when the disease is actually present
and in 10% of all cases when it is not. If a result is positive, the test is repeated. Assume that the second
test is independent of the first test. If the prevalence of the disease is 1 in 50, and an individual tests
positive twice, what is the probability that the individual actually has the disease?
Solution