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 

            (ii) not more than 2 will bear 'Y'  mark.

Answers (1)
S seema garhwal

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 replacementt.

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

Z has binomail distribution with n=6.

 P(Z=z)=^nC_Z P^{n-Z}q^Z

P(not\, more\, than\, 2\, bear\, Y) =P(Z\leq 2)

                       =P(Z=0)+P(Z=1)+P(Z=2)

                      =^6C_0 (\frac{2}{5})^{6} (\frac{3}{5})^0+^6C_1 (\frac{2}{5})^{5} (\frac{3}{5})^1+^6C_2 (\frac{2}{5})^{4} (\frac{3}{5})^2

                     = (\frac{2}{5})^{6} +6 (\frac{2}{5})^{5} (\frac{3}{5})^1+15 (\frac{2}{5})^{4} (\frac{3}{5})^2

                  = (\frac{2}{5})^{4} [(\frac{2}{5})^{2}+6 (\frac{2}{5}) (\frac{3}{5})+15(\frac{3}{5})^2]

               = (\frac{2}{5})^{4} [\frac{4}{25}+\frac{36}{25}+\frac{135}{25}]

                  = (\frac{2}{5})^{4} [\frac{175}{25}]

                  = (\frac{2}{5})^{4} [7]

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