Question

A box contains 100 balls, of which r are red. Suppose that the balls are drawn from the box one at a time, at random without replacement. Determine (a) the probability that the first ball drawn will be red; (b) the probability that the 50th ball drawn will be red; and (c) the probability that the last ball drawn will be red.

Answer 1

Answer:

(a) $\frac{r}{100}$

(b) $\frac{r}{100}$

(c) $\frac{r}{100}$

Step-by-step explanation:

Given,

The total number of balls = 100,

Red balls = r

So, the remaining balls = 100 - r,

(a) ∵ The probability that first ball drawn will be red

$=\frac{\text{Red balls}}{\text{Total balls}}$

$=\frac{r}{100}$

(b) Also, the probability of a ball other than red ball = $1-\frac{r}{100}$

$=\frac{100-r}{100}$

So, the probability of getting red ball in second thrawn( one is red second is red or one is not red second is red ),

$=\frac{r}{100}\times \frac{r-1}{99}+\frac{100-r}{100}\times \frac{r}{99}$

$=\frac{r}{99}[\frac{r-1}{100}+\frac{100-r}{100}]$

$=\frac{r}{99}[\frac{r-1+100-r}{100}]$

$=\frac{r}{99}[\frac{99}{100}]$

$=\frac{r}{100}$

Now, the the probability of getting red ball in third thrawn,

$=\frac{r}{100}\times \frac{r-1}{99}\times \frac{r-2}{98}+\frac{100-r}{100}\times \frac{r}{99}\times \frac{r-1}{98}+\frac{100-r}{100}\times \frac{99-r}{99}\times \frac{r}{98}$

$=\frac{r}{100}$

......so on,...

This pattern will be followed in every trials,

Hence, the probability that the 50th ball drawn will be red = $\frac{r}{100}$

(c) Similarly,  the probability that the last ball drawn will be red = $\frac{r}{100}$