Question

Two defective tubes get mixed up with 2 good ones. The tubes are tested, one by one,

until both are found. What is the probability that the last defective tube is obtained on

(i) the second test, (ii)the third test and (iii) the fourth test​

Answer 1

Answer: here is what I got

Step-by-step explanation:

Answer 2

Concept:

By dividing the favourable number of possibilities by the entire number of potential outcomes, the probability of an occurrence may be estimated using the probability formula.

Given:

There are 2 defective tubes and 2 good tubes.

Find:

The probability of the last defective tube in the second test, third test, and fourth test.

Solution:

Total number of tubes are = 4 ( 2 good + 2 defective).

The total arrangements of tubes = $\frac{4!}{2!2!}$.

                                                        = 6.

Take defective tubes as X and good tubes as Y.

The number of arrangements are,

XXYY, YYXX, XYXY, XYYX, YXXY, YXYX.

(i) The last defective tube occurring in the second test is in XXYY case only.

Hence, the probability of the last defective tube occurring in the second test is $\frac{1}{6}$.

(ii) The last defective tube occurring in the third test is in XYXY and YXXY cases only.

The probability of the last defective tube occurring in the third test is =  $\frac{2}{6} = \frac{1}{3}$.

Hence, the probability of the last defective tube occurring in the third test is $\frac{1}{3}$.

(iii) The last defective tube occurring in the third test is in YYXX, XYYX and YXYX cases only.

The probability of the last defective tube occurring in the fourth test is =  $\frac{3}{6} = \frac{1}{2}$.

Hence, the probability of the last defective tube occurring in the third test is $\frac{1}{2}$.