Chapter 15.4, Problem 55E

To determine

To find: the probability that the four odd numbers are drawn.

Answer to Problem 55E

  $P\left(\text{four odd numbers}\right)=\frac{253}{4006}$ .

Explanation of Solution

Given:

50 tickets, numbered consecutively from 1 to 50 are placed in the box.

Four tickets are drawn without replacement.

Calculation:

As there are 25 odd numbers between 1 and 15, because 5 odd number set is between only 1 to 10.

Similarly, another 5 sets are between only 11 to 20 and so on.

since the tickets are drawn without replacement, with each consecutive draw, there is 1 fewer odd number and 1 fewer ticket.

Therefore, the probability that the four odd numbers are drawn in the box is:

  $\begin{array}{l}P\left(\begin{array}{l}\text{four odd numbers }\\ \text{without replacement}\end{array}\right)=\frac{25}{50}\times \frac{24}{49}\times \frac{23}{48}\times \frac{22}{47}\times \frac{21}{46}\\ =\frac{253}{4006}\end{array}$

Hence, probability that the four odd numbers are drawn in the box is $P\left(\text{four odd numbers}\right)=\frac{253}{4006}$