CHAPTER 7

Copyright ©2001 The McGraw-Hill Companies.
Any use is subject to the Terms of Use and Privacy Policy.
McGraw-Hill Higher Education is one of the many fine businesses of the The McGraw-Hill Companies.

CHAPTER 7

SOLUTIONS TO PROBLEMS

(a).

Since these are mutually exclusive events, P(NY or CA or TX) = P(NY) + P(CA) + P(TX) =

(b).

(c). Since these are mutually exclusive events, P(Northeast or West) = P(Northeast) + P(West) =

(d).

Given the 136,046 prisoners in the Midwest, what is the probability that the prisoner is from Michigan?

These two probabilities are different because the second probability is a conditional probability so the denominator is the total number incarcerated in the midwest. Michigan prisoners make up 5% of the total prisoner population, but 23% of the prisoners in the midwest.

Since these are mutually exclusive events, P(Illinois, Ohio or Indiana½Midwest) = P(Illinois) + P(Ohio) + P(Indiana):

(e). P(New York½Northeast) =

P(New Jersey or Pennsylvania½Northeast) =

(f). P(Maine & Louisiana) = 0, because these are mutually exclusive events. A prisoner cannot at the same time be incarcerated both in Maine and Louisiana.
(a). Since there are 4 Aces in a deck of 52 cards,

(b). P(King of Spades) = .

(c). These are not mutually exclusive events since a heart can also be a seven (the seven of hearts). Therefore, P(Heart or Seven) = P(Heart) + P(Seven) - P(Seven of Hearts) =

(d). This question asks for the probability of three events, a King, a two, or a Club. These events are not mutually exclusive since both a King and a two may also be Clubs (the King of Clubs and the two of Clubs). Therefore, P(King or Two o r Clubs) = P(King) + P(Two) + P(Clubs) - P(King of Clubs) - P(Two of Clubs) =

(e). Not counting the Ace or face cards, the odd diamonds are the three, five, seven, and nine of diamonds. Since these are mutually exclusive events, P(Three of Diamonds or Five of Diamonds or Seven of Diamonds or Nine of Diamonds) = P(Thre e of Diamonds) + P(Five of Diamonds) + P(Seven of Diamonds) + P(Nine of Diamonds) = .

(f). These are mutually exclusive events since one card cannot be both a Heart and a Spade. Therefore P(Heart & Spade) = 0.
(a). The probability of getting no heads is the probability of getting six tails in six flips of a coin. P(T T T T T T) = (.5)(.5)(.5)(.5)(.5)(.5) = (.5)⁶ = .0156.

(b). This problem has a few parts to it. The probability of getting one head or one tail is equal to P(H T T T T T) + P(T H H H H H). Let's first find the first of these probabilities. P(H T T T T T) = (.5)(.5)(.5)(.5)(.5)(.5) = (.5)⁶ = .0156. But, this is the probability of getting one head followed by five tails exactly in that order. Ther e are other ways of getting one head and five tails, such as T H T T T T. Exactly how many ways are there to get one head and five tails from six flips of a coin? Using our combination rule we can determine this to be: . There are, then, six ways to order five tails and one head. The probability of obtaining one head and five tails in any order, therefore, is equal to 6(.0156) or .0936.

Now we have to find the probability of getting one tail and five heads P(T H H H H H). Fortunately, this is the same as the probability of one head and five tails that we just figured out above. This probability is equal to 6(.0156) or .093 6.

Using our addition rule, the probability of getting one head in six flips of a coin or one tail in six flips of a coin is equal to .0936 + .0936 = .1872.

(c). To find the probability of three tails in six flips of a coin, let's first find that probability for any ordering of heads and tails. P(T T T H H H) is equal to (.5)(.5)(.5)(.5)(.5)(.5) = (.5)⁶ = .0156. Keep in mind that this is equal to the probability of getting three tails followed by three heads. We now have to determine how many ways we can order three tails and three heads without regard to order. This is a combination problem again, and is equal to . There are twenty ways to order three tails and three heads. The probability of getting three tails and three heads then is equal to 20(.0156) or .312.

(d). The probability of getting four or fewer heads in six flips of a coin is the probability of getting four heads, or three heads, or two heads, or one head, or no heads. Written below it is:
```
P(H H H H T T) +
P(H H H T T T) +
P(H H T T T T) +
P(H T T T T T) +
P(T T T T T T)
```
We have to find each of these probabilities separately, then add them. Let's work them in order.

P(H H H H T T): first what is the probability of getting four heads and two tails in any order? It is (.5)⁶ or .0156. Now, how many ways of ordering four heads and two tails are there? This is a combination problem, . The probability of four heads and two tails in any order, then, is equal to 15(.0156) or .234.

P(H H H T T T): the probability of three heads and three tails in this exact order is (.5)⁶ or .0156, and there are ways to order these events. The probability of three heads and three tails is equal t o 20(.0156) or .312.

P(H H T T T T): the probability of two heads and four tails in this exact order is (.5)⁶ or .0156, and there are ways to order these events. The probability of two heads and four tails is equal to 15(. 0156) or .234. Notice that this probability is the same as the probability of four heads and two tails.

P(H T T T T T): the probability of one head and five tails in this exact order is (.5)⁶ or .0156, and there are ways to order these events. The probability of one head and five tails is equal to 6(.015 6) or .094.

P(T T T T T T): the probability of six tails is equal to (.5)⁶ or .0156, and this is the only possible ordering of these events.

The probability of four or fewer heads, then, is equal to .234 + .312 + .234 + .094 + .0156 = .89
The full probability distribution of 4 flips of a coin is as follows:
```
P(T T T T) = (.5)⁴ = .0625

P(H T T T) =     

P(H H T T) =     

P(H H H T) =     

P(H H H H) = (.5)⁴ = .0625
```
The most likely outcome of four flips of a coin is two heads and two tails [P(HHTT) = .375]. This is because if the coin is fair the probability of a head is the same as the probability of a tail (.5). The least likely outcomes are no heads and no tails. No, you would not reject the null hypothesis that the coin is fair. With a fair coin, where the probability of a head is .5, the probability that you would get four tails from four flips is .0625. This is greater than the .05 level of signi ficance you choose. The outcome would have to have a probability less than .05 in order for you to reject the null hypothesis.
(a). Since the probability of being white is the complement of the probability of being black, you could also have found P(White) = 1 - P(Black) = 1 - .327 = .673.

(b).

(c). P(Black & Property Crime) = P(Black) * P(Property Crime½Black) = . The conditional probability is used here because being black and committing a property crime are not stat istically independent events. P(Black) = .327 but P(Black½Property Crime) = .28. Since P(A) ¹ P(A½B), the events are not independent.

(d). P(White & Drug Crime) = P(White) * P(Drug Crime½White) = .

(e). P(Drug Crime & Property Crime) = 0. These events are mutually exclusive because there was only one offense charged for each youth. A given youth could not, therefore, be charged with two different crimes.

(f). P(White & Black) = 0. These are mutually exclusive events.

(g). P(White & Property Crime) = P(White) * P(Property Crime ½White) = .

(h). P(Person Crime) = .

(i). P(Black & Person Crime) = P(Black) * P(Person Crime½Black) = .

(j). P(White & Person Crime) = P(White) * P(Person Crime½White) = .
(a). P(Severe Injury) + P(Minor Injury) = .

(b). P(Violent Criminal) + P(Burglar) = .

(c). P(Burglar & Severe Injury) Before we answer this, let's see if the target of the gun use (violent v. burglary v. family member) and the consequence (severe v. minor v. no injury) are independent events. Is the probability of using i t against a burglar, for example, the same as the conditional probability of using it against a burglar given that it involved serious injury? P(Burglar) = , the conditional probability P(Burglar½Serious Injury) is . The two are not equal, so the events are not independent. The P(Burglar & Severe Injury) = P(Burglar) * P(Severe Injury½Burglar) = .

(d). P(Family Member & Severe Injury) = P(Family Member) * P(Severe Injury½Family Member) = .

Return to Table of Contents Return to Criminal Justice Home Page