Question

A bag contains 6 red, 4 white and 8 blue balls. If 3 balls are drawn at randomfrom the bag, find the probability that the balls are of different colours.

Answer 1

Hi there!
Here's the answer:

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¶¶ Points to remember ¶¶


¶ Probability of an Event
= No. of Favourable outcomes for the Event / Total No. of Outcomes.

¶ Sum of Probabilities = 1
=> P(S) = 1

¶ Probability(Drawing same coloured balls) + P(Drawing different coloured balls) = 1
=> P(E) + P(E') = 1
=> P(E) = 1 - P(E')

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¶¶ SOLUTION:

In a bag,
No. of Red Balls = 6
No. of White Balls = 4
No. of Blue Balls = 8
Total No. of balls = 18

Let S be sample space
n(S) - no. of ways of drawing 3 balls from 18 balls
n(S) = 18C3 = (18×17×16)/(3×2×1) = 816


Let E be Event of drawing 3 different balls from the bag

E' is the Event of drawing 3 balls of the same colour
Then, E' = event of drawing (3 ball out of 6) or (3 ball out of 4) or (3 ball out of 8)

No. of possible cases for Event E' is given by
n(E') = 6C3 + 4C3 + 8C3
=> n(E') = 20 + 4 + 56
=> n(E') = 80

Probability of Drawing 3 balls of same colour
P(E') = n(E') / n(S)
=> P(E') = 80 / 816 = 5 / 51.

°•° P(E) + P(E') = 1
=> P(E) = 1 - P(E')
=> P(E) = 1 - (5/51)
=> P(E) = 46/51.


•°• Required Probability = 46/51


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Hope it helps