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  • An urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15 bear a mark 'Y' (i) all will bear X mark

Q. 5   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 down and it is replaced. If 6 balls are drawn in this way, find the probability that

            (i) all will bear 'X' mark.

Answers (1)

best_answer

Total balls in urn = 25

Balls bearing mark 'X' =10

Balls bearing mark 'Y' =15

P(ball\, \, bearing\, mark\, 'X')=\frac{10}{25}=\frac{2}{5}

P(ball\, \, bearing\, mark\, 'Y')=\frac{15}{25}=\frac{3}{5}

 6 balls are drawn with replacement.

Let Z be a random variable that represents a number of balls with Y mark on them in the trial.

Z has a binomial distribution with n=6.

 P(Z=z)=^nC_Z P^{n-Z}q^Z

P(Z=0)=^6C_0 (\frac{2}{5})^{6} \frac{3}{5}^0

  P(Z=0)=^6C_0 (\frac{2}{5})^{6}

Posted by

seema garhwal

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