Math 2000 - Solutions 9

Spring 2001

Unit 8C

4b. Each coin toss is independent and the probability of a tail on a single toss is 1/2. So the probability of ten consecutive tails is P(10T) = (1/2)¹⁰ = 1/1024 = 0.00098.

4c. The outcomes of lottery tickets are independent. So with a 1/8 probability of winning with a single ticket, the probability of winning on four tickets in a row is P(4 wins) = (1/8)⁴ = 1/4096 = 0.0002.

6. The probability of drawing a red marble on the first try is 20/30 = 2/3 = 0.67. So the probability of choosing a red marble twice in a row with replacement is 2/3 ´ 2/3 = 4/9 = 0.444. However, if the first marble is not replaced, then the outcome of the second draw depends on the outcome of the first. If a red marble is drawn the first time, this leaves 19 red marbles in the bag out of a total of 29 marbles . So the probability of choosing a red marble on the second draw given that a red marble was chosen on the first draw is 19/29. So we have P(two red marbles) = P(red marble on first draw) ´ P(red marble on second draw given a red marble on first draw) = 2/3 ´ 19/29 = 0.437 ¾ slightly less than the replacement with replacement.

8c. Each time a jack is drawn from the deck, the number of jacks decreases by one and the number of cards in the deck decreases by one. Therefore,

The chances of drawing four jacks in a row are about four in a million.

10d. d. In this problem, we are dealing with non-mutually-exclusive events since choosing a woman does not exclude choosing an American. We can reason in two ways here. (i) The only way not to choose a woman or an American is to choose an English man. The probability of choosing an English man is 1/4, so the chance of not choosing a woman or an American is 1/4. This means the probability of choosing a woman or an American is 3/4.

Alternatively, we can use the formula for non-mutually-exclusive events which give us

P(woman or American) = P(woman) + P(American) - P(woman and American) = 1/2 + 1/2 - 1/4 = 3/4.

Either way, the probability of choosing a woman or an American from the group is 3/4 or 0.75.

14b. The probability of flipping a tail is P(T) = 1/2. The probability of not flipping a tail is P(no T) = 1 - P(T) = 1/2. The probability of not flipping a tail in 4 flips (independent events) is

P(no T in 4 flips) = P(no T)⁴ = (1/2)⁴ = 1/16.

Therefore, the probability of flipping at least one tail in 4 rolls is P(at least one T) = 1 - 1/16 = 15/16.

18. a. We can use the “at least once” rule to analyze the probability of an encounter with an HIV-carrying person. Suppose that 2% is the percentage of carriers of HIV in the general population, which gives the probability of an encounter, not the probability of infection. Note that six partners in each of four years means a total of 24 partners. Therefore,

The probability of an encounter with a person carrying HIV assuming 24 partners is better than one in three.

b. If we assume the same incidence rate (2%), but now assume that there are 4 ´ 12 = 48 encounters, we have that

These results may defy intuition. First of all, the chances are higher than most people might expect. But notice further that doubling the number of partners (24 to 48) almost doubles the chances of a risky encounter (0.38 to 0.62).

Lottery Problem

Recall that to count the total outcomes for a lottery, we need to use combinations. The total number of outcomes for the three lotteries are as follows:

Lottery 1:

Lottery 2:

Lottery 3:

Lottery 1 is the best bet; because it has the fewest outcomes, your chances of matching the winning combination are greatest.