83.1k views
0 votes
a box contains n balls of which 3 are red and the rest are white. 2 balls are selected at random, one after the other without replacement. If the probability of selecting 2 red balls is 3/28. Find the value of n. the probability that one is red and the other is white

1 Answer

1 vote

To find the value of n, set up an equation using the given information and solve for n. The value of n is 93. The probability that one ball is red and the other is white is 3(n-3)/(n(n-1)).

To find the value of n, we can set up an equation using the given information. The probability of selecting 2 red balls is 3/28. This means that the first ball selected has a probability of 3/n (where n is the total number of balls) of being red, and the second ball selected has a probability of 2/(n-1) of being red, since one red ball has already been selected. Multiplying these probabilities together gives us (3/n) * (2/(n-1)) = 3/28.

To solve for n, we can cross-multiply and solve the resulting quadratic equation. After simplifying, we get 6n - 9 = 84. Solving for n gives us n = 93.

The probability that one ball is red and the other is white can be found by subtracting the probability of selecting 2 red balls from 1. Since there are 3 red balls and (n-3) white balls, the probability is (3/n) * ((n-3)/(n-1)), which simplifies to 3(n-3)/(n(n-1)).

User Fernando Barbeiro
by
7.9k points