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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) At least one ball will bear ‘ ‘ mark. (ii) The number of balls with ‘ ‘ mark and ‘ ‘ mark will be equal.
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) At least one ball will bear ‘ \mathrm{Y} ‘ mark.
(ii) The number of balls with ‘ \mathrm{X} ‘ mark and ‘ \mathrm{Y} ‘ mark will be equal.
CBSE | Class 12 | 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) At least one ball will bear ‘ \mathrm{Y} ‘ mark.
(ii) The number of balls with ‘ \mathrm{X} ‘ mark and ‘ \mathrm{Y} ‘ mark will be equal.

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.

(i) Now, Probability (At least one ball will bear ‘ \mathrm{Y} ‘ mark) = \mathrm{P}(\mathrm{Z} \geq 1)

=1-\mathrm{P}(\mathrm{Z}=0)

=1-\left(\frac{2}{5}\right)^{6}

(ii) Probability (Having equal number of balls with ‘ \mathrm{X}^{\prime} mark and ‘ \mathrm{Y}^{\prime} mark) =\mathrm{P}(\mathrm{Z}

=3

={ }^{6} C_{3}\left(\frac{2}{54}\right)^{3}\left(\frac{3}{5}\right)^{3}

=\frac{20 \times 8 \times 27}{15625}

=\frac{864}{3125}

 

 

i

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