Question

professor maxwell
name____________________
intro to statistics
quiz 4
spring 2015
date________

1. a card is selected at random from a standard deck of 52 playing cards. find each probability.

(a) randomly selecting a club or a 3
(b) randomly selecting a red suit or a king
(c) randomly selecting a 9 or a face card

1. the table shows the results of a survey that asked 3203 people whether they wore contacts or glasses. a person is selected at random from the sample. find the probability of each event.

	only contacts	only glasses	both	neither	total
male	64	841	177	456	1538
female	189	427	368	681	1665
total	253	1268	545	1137	3203

(a) the person wears only contacts or only glasses.
(b) the person is male or wears both contacts and glasses.
(c) the person is female or wears neither contacts nor glasses.
(d) the person is male or does not wear glasses.

Explanation:

Step1: Recall probability formula

The probability of an event $A$ or $B$, $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.

Step2: Solve 11(a)

There are 13 clubs in a deck, so $P(\text{club})=\frac{13}{52}$. There are 4 cards with number 3, so $P(3)=\frac{4}{52}$. The card that is both a club and a 3 is 1 card, so $P(\text{club}\cap3)=\frac{1}{52}$. Then $P(\text{club}\cup3)=\frac{13 + 4-1}{52}=\frac{16}{52}=\frac{4}{13}$.

Step3: Solve 11(b)

There are 26 red - suited cards, so $P(\text{red suit})=\frac{26}{52}$. There are 4 kings, so $P(\text{king})=\frac{4}{52}$. There are 2 red kings, so $P(\text{red suit}\cap\text{king})=\frac{2}{52}$. Then $P(\text{red suit}\cup\text{king})=\frac{26 + 4-2}{52}=\frac{28}{52}=\frac{7}{13}$.

Step4: Solve 11(c)

There are 4 cards with number 9, so $P(9)=\frac{4}{52}$. There are 12 face - cards, so $P(\text{face card})=\frac{12}{52}$. There are no cards that are both 9 and a face - card, so $P(9\cap\text{face card}) = 0$. Then $P(9\cup\text{face card})=\frac{4+12 - 0}{52}=\frac{16}{52}=\frac{4}{13}$.

Step5: Solve 12(a)

The number of people who wear only contacts is 253 and only glasses is 1268. So $P(\text{only contacts}\cup\text{only glasses})=\frac{253 + 1268}{3203}=\frac{1521}{3203}$.

Step6: Solve 12(b)

The number of males is 1538 and the number of people who wear both is 545. The number of males who wear both is 177. So $P(\text{male}\cup\text{both})=\frac{1538+545 - 177}{3203}=\frac{1906}{3203}$.

Step7: Solve 12(c)

The number of females is 1665 and the number of people who wear neither is 1137. The number of females who wear neither is 681. So $P(\text{female}\cup\text{neither})=\frac{1665+1137 - 681}{3203}=\frac{2121}{3203}$.

Step8: Solve 12(d)

The number of males is 1538. The number of people who do not wear glasses is $253+456 + 681=1390$. The number of males who do not wear glasses is $64 + 456=520$. So $P(\text{male}\cup\text{no glasses})=\frac{1538+1390 - 520}{3203}=\frac{2408}{3203}$.

Answer:

11(a): $\frac{4}{13}$
11(b): $\frac{7}{13}$
11(c): $\frac{4}{13}$
12(a): $\frac{1521}{3203}$
12(b): $\frac{1906}{3203}$
12(c): $\frac{2121}{3203}$
12(d): $\frac{2408}{3203}$