Question:
An urn contains 25 balls of which 10 balls bear a mark X and the remaining 15 bear a mark Y. A ball is drawn at random from the urn, its mark is noted downand it is replaced. If 6 balls are drawn in this way, find the probability that (i) all will bear X mark. (ii) not more than 2 will bear Y mark. (iii) at least one ball will bear Y mark. (iv) the number of balls with X mark and Y mark will be equal.
Answer:
Total number of balls = 25
Balls bear mark X = 10
Balls bear mark Y = 15
Probability that balls bear mark X = p = 10/25 = 2/5
Probability that balls bear mark Y =q = 15/25 = 3/5
Now 6 balls are drawn with replacement. So number of trilas are bernoulli trilas.
Let Z is the random variable that represents the number of balls with X mark on them in the trial.
So Z has the binomial distribution with n = 6 and p = 2/5
So P(Z = z) = nCz * pn-z * qz
1. Probability that all balls are bear mark X = P(Z = 0) = 6C0 * (2/5)6-0 * (3/5)0
= (2/5)6
2. Probabiliy that not more than 2 balls bear mark Y = P(Z<=2) = P(Z=0) + P(Z=1) + P(Z=2)
= 6C0 * (2/5)6 * (3/5)0 + 6C1 * (2/5)6-1 * (3/5)1 + 6C2 * (2/5)6-2 * (3/5)2
= (2/5)6 + 6*(2/5)5 *(3/5) + 15* (2/5)4 * (3/5)2
= (2/5)4 *{2/5 + 36/25 + 135/25}
= (2/5)4 *{175/25}
= 7(2/5)4
3. Probability that at least one ball bear mark Y = P(Z>=1) = 1 - P(Z = 0)
= 1- (2/5)6
4 . Probability that equal number of balls bear with mark X and Y = P(Z = 3)
= 6C3 * (2/5)3 * (3/5)3
= (6*5*4)/(3*2) * 8/125 * 27/125
= (5*4) * 8/125 * 27/125
= 5* 8/25 * 27/125
= 864/3125
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