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Question

An urn contains 25 balls of which 10 bear a mark 'X' and the remaining 15 bear a mark 'Y'. A ball is drawn at random from the urn, its mark noted down and it is replaced. If 6 balls are drawn in this way, find the probability that
(i) All will bear mark 'X'
(ii) Not more than two will bear 'Y' mark

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Solution

From the given statement :
Total number of balls = 25
Balls bear mark Y=15
Balls bear mark X=10
Probability that balls bear mark X=p=1025=25

Probability that balls bear mark Y=q=1525=35

Now 6 balls are drawn with replacement

So, the number of trials are Bernoulli trials.

Let Z is the random variable that represents the number of balls with X mark on them in the trial.

Hence, Z has the binomial distribution with n=6,p=25

P(Z=z)=ncz×pnz×qz

1.Probability that all balls bear mark X=P(Z=0)=6c0×(25)60×(35)0

= (25)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×(25)60×(35)0+6c1×(25)61×(35)1+6c2×(25)62×(35)2

(25)6+6×(25)5×35+15×(25)4×(35)2

= (25)4×17525

= 7(25)4


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