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Home » An urn contains 25 balls of which 10 balls bear a mark ‘ ‘ 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 ‘ ‘ mark. (ii) Not more than 2 will bear ‘Y’ mark.
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.
(ii) Not more than 2 will bear ‘Y’ mark.
CBSE | Maths | NCERT | Probability
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.
(ii) Not more than 2 will bear ‘Y’ mark.

Solution:
(i) Given in the question that,

Total number of balls in the urn =25

Number of balls bearing mark { }^{\prime} \mathrm{X}^{\prime}=10

Number of balls bearing mark { }^{\prime} \mathrm{y}^{\prime}=15

In this equation, p represents the probability of balls bearing mark’X ‘, while q represents the probability of balls not bearing mark’X ” ‘ \mathrm{Y}^{\prime}

p=10 / 25=2 / 5 and q=15 / 25=3 / 5

Six balls are drawn with replacement at this point. As a result, the number of trials is represented by a Bernoulli triangle.

Assume, \mathrm{Z} be the random variable that represents the number of balls bearing ‘ \mathrm{Y} ‘ mark in the trials

\therefore Z has a binomial distribution where n=6 and p=2 / 5

P(Z=z)={ }^{n} C_{z} p^{n-z} q^{2}

Hence, \mathrm{P} (All balls will bear mark ‘ \left.\mathrm{X}^{\prime}\right)=\mathrm{P}(\mathrm{Z}=0 )

={ }^{6} C_{0}\left(\frac{2}{5}\right)^{6}
=\left(\frac{2}{5}\right)^{6}

(ii) Probability (Not more than 2 will bear ‘ \mathrm{Y} ‘ mark) =\mathrm{P}(\mathrm{Z} \leq 2)

=\mathrm{P}(\mathrm{Z}=0)+\mathrm{P}(\mathrm{Z}=1)+\mathrm{P}(\mathrm{Z}=2)

={ }^{6} C_{0}(p)^{6}(q)^{0}+{ }^{6} C_{1}(p)^{5}(q)^{1}+{ }^{6} C_{2}(p)^{4}(q)^{2}

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

=\left(\frac{2}{5}\right)^{4}\left[\left(\frac{2}{5}\right)^{2}+6\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)+15\left(\frac{3}{5}\right)^{2}\right]
=\left(\frac{2}{5}\right)^{4}\left[\frac{175}{25}\right]
=7\left(\frac{2}{5}\right)^{4}

i

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