Question

In a previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data were compiled into the following table.

For the following, suppose that you randomly select one player from the 49ers or Cowboys.

a. Find the probability that his shirt number is from 1 to 33.

b. Find the probability that he weighs at most 210 pounds.

c. Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds.

d. Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds.

e. Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210 pounds.

Short answer

a. $\frac{26}{106}$

b. $\frac{33}{106}$

c. $\frac{21}{106}$

d. $\frac{38}{106}$

e. $\frac{21}{33}$

Step-by-step solution

Given information

The factual data were compiled into the following table.

Shirt#	≤210	211–250	251–290	>290	Total
1–33	21	5	0	0	26
34–66	6	18	7	4	35
66–99	6	12	22	5	45
Total	33	35	29	9	106

Part (a) Step 1: Explanation 

the probability that his shirt number is from 1 to 33:

$P\left(1to33\right)=\frac{26}{106}$

Part (b) Step 1: Explanation 

the probability that he weighs at most 210 pounds:

$P(\le 210)=\frac{33}{106}$

Part (c) Step 1: Explanation 

the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds:$P(1to33AND\le 210)=\frac{21}{106}$

Part (d) Step 1: Explanation 

the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds:

$$\begin{gathered} P(1to33OR\le 210)=P\left(1to33\right)+P(\le 210)-P(1to33AND\le 210) \\ P(1to33OR\le 210)=\frac{26}{106}+\frac{33}{106}-\frac{21}{106}=\frac{38}{106} \end{gathered}$$

Part (e) Step 1: Explanation 

the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210 pounds.

$P(\overline{)1to33}\le 210)=\frac{21}{33}$