CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Probability

Q.11. A car manufacturing factory has two plants X and Y. Plant X manufactures 70% of the cars and plant Y manufactures 30%. At plant X, 80% of the cars are rated of standard quality and at plant Y, 90% are rated of standard quality. A car is picked up at random and is found to be of standard quality. Find the probability that it has come from plant X.

Solution :

Let E1 and E2 be the event that car is manufactured by plant X and Y respectively. Let E be the event that the car is of standard quality.
P(E1) = 70/100 = 7/10; P(E2) = 30/100 = 3/10; P(E/E1) = 80/100 = 8/10; P(E/E2) = 90/100 = 9/10.
Now, P(E1/E) = [P(E1)×P(E/E1)]/[P(E1)×P(E/E1) + P(E2)×P(E/E2)]
= [7/10×8/10]/[7/10×8/10 + 3/10×9/10]
= (7×8)/(7×8 + 3×9)
= 56/83. [Ans.]

Q12. A doctor is to visit a patient. From the past experience it is known that the probabilities of the doctor coming by train, bus, scooter or taxi are 1/10, 1/5, 3/10 and 2/5 respectively. The probabilities that he will be late are 1/4, 1/3, and 1/12 if he comes by train, bus or scooter respectively but by taxi he will not be late. When he arrives he is late. What is the probability that he came by bus ?

Solution :

Let the events that the doctor comes by train, bus, scooter and taxi respectively be A, B, C and D.
Therefore, P(A) = 1/10, P(B) = 1/5, P(C) = 3/10 and P(D) = 2/5.
Let E be the event that the doctor comes late
Therefore, P(E/A) = 1/4, P(E/B) = 1/3, P(E/C) = 1/2 and P(E/D) = 0.
Probability that doctor comes late by bus = P(B/E)
= [P(B)P(E/B)]/[P(A)P(E/A) + P(B)P(E/B) + P(C)P(E/C) + P(D)P(E/D)]
= [1/5×1/3]/[1/10×1/4 + 1/5×1/3 + 3/10×1/12 + 2/5×0]
= [1/15]/[1/40 + 1/15 + 3/120]
= [1/15]/[(3 + 8 + 3)/120]
= (1/15)(120/14)
= 4/7.

[Ans.]

13.4. Probability Distribution.

Q.1. Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of number of jacks.

Solution :

Let X be the event of drawing jack from 52 playing cards in two draws.
Here, X = 0, 1, 2.
Therefore, P(X = 0) = P(Not a jack in 1st draw and not a jack in 2nd draw)
= P(not a jack).P(not a jack)
= (48/52)(48/52)
= 144/169.
P(X = 1) = P(Jack & not a jack or Not a jack & jack)
=P(jack & not a jack) + P(not a jack & jack)
= P(jack).P(not a jack) + P(not a jack).P(jack)
= (4/52)(48/52) + (48/52)(4/52)
= 24/169.
P(X = 2) = P(Jack & Jack) = P(Jack).P(Jack) = (4/52)(4/52) = 1/169.
Hence required probability distribution is :

Paper By Mr. M.P.Keshari
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