NCERT Class XII Mathematics Chapter - - Solutions
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Question : 87
Total: 101
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.
(iii) at least one ball will bear ‘Y’ mark.
(iv) the number of balls with ‘X’ mark and ‘Y’ mark will be equal.
(i) all will bear ‘X’ mark.
(ii) not more than 2 will bear ‘Y’ mark.
(iii) at least one ball will bear ‘Y’ mark.
(iv) the number of balls with ‘X’ mark and ‘Y’ mark will be equal.
Solution:
p be the probability of drawing a ball with mark X =
=
q be the probability of drawing a ball with mark Y=
=
Let X has a binomial distributionn = 6 , p =
, q =
∴ P ( X = r ) = n C r ( q ) n − r p r
(i) P (all balls bear ‘X’ mark)= P ( x = 6 ) = 6 C 6 q 0 p 6 = 1 × 1 × (
) 6 = (
) 6
(ii) P(Not more than 2 balls bear ‘Y’ mark)
= P(at least 4 balls bear ‘X’ mark)
∴ Required probability = P(X = 4) + P(X = 5) + P(X = 6)= 6 C 4 q 2 p 4 + 6 C 5 q p 5 + 6 C 6 q 0 p 6 = 15 (
) 2 (
) 4 + 6 (
) (
) 5 + ( 1 ) ( 1 ) (
) 6 = 7 (
) 4
(iii) P (at least one ball will bear ‘Y’ mark) = P (at most 5 balls will bear ‘X’ mark)
Required probability = P(X ≤ 5) = 1 – P(X = 6)
= 1 − 6 C 6 q 0 p 6
= 1 − ( 1 ) ( 1 ) (
) 6
= 1 − (
) 6
(iv) P (No. of balls with ‘X’ mark) = P(No. of balls with ‘Y’ mark)
⇒ P ( X = 3 ) = P ( Y = 3 ) = 6 C 3 q 3 p 3
= 20 (
) 3 (
) 3 =
q be the probability of drawing a ball with mark Y
Let X has a binomial distribution
(i) P (all balls bear ‘X’ mark)
(ii) P(Not more than 2 balls bear ‘Y’ mark)
= P(at least 4 balls bear ‘X’ mark)
∴ Required probability = P(X = 4) + P(X = 5) + P(X = 6)
(iii) P (at least one ball will bear ‘Y’ mark) = P (at most 5 balls will bear ‘X’ mark)
Required probability = P(X ≤ 5) = 1 – P(X = 6)
(iv) P (No. of balls with ‘X’ mark) = P(No. of balls with ‘Y’ mark)
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