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

Probability

Q.3. 12 cards numbered 1 to 12, are placed in a box, mixed up thoroughly and then a card is drawn at random from the box. If it is known that the number on the drawn card is more than 3, find the probability that it is an even number.

Solution :

We have, S = {1, 2, 3, 4, ----------, 12}.
Total events = n(S) = 12.
Let E : number on the drawn card is more than 3,
F : number on the drawn card is even number.
E = {4, 5, 6, 7, 8, 9, 10, 11, 12}
F = {2, 4, 6, 8, 10, 12}
E∩F = {4, 6, 8, 10, 12}
P(E) = 9/12, P(F) = 6/12, P(E∩F) = 5/12
P(F/E) = P(E∩F)/P(E) = (5/12)/(9/12) = 5/9.[Ans.]

Q.4. A family has two children. What is the prtobability that both the children are boys, given that at least one of them is a boy ?

Solution :

Let B be the boy and G be the girl. Let S be the sample space of the experiment. Then S = {(B, B), (B, G), (G, B), (G, G)}.
And n(S) = 4.
Let the event E : both the children are boys
And F : at least one of the child is a boy.
Therefore, E = {(B, B)} and F = {(B, B), (B, G), (G, B)}and E∩F {(B, B)}.
P(E) = 1/4 , P(F) = 3/4 , P(E∩F) = 1/4 .
Therefore, P(Both the children are boys) = P(E/F)
= P(E∩F)/P(F)
= (1/4)/(3/4)
= 1/4×4/3
= 1/3. [Ans.]

13.3. Bayes’ Theorem.

Q.1. In a bolt factory, machines A, B, C manufacture 25%, 35% and 40% respectively of the total bolts. Of their output 5%, 4% and 2% respectively are defective bolts. A bolt is drawn at random and is found to be defective. Find the probability that it is manufactured by machine B.

Solution :

Let E be the probability of drawing defective bolt. Let E1, E2 and E3 be the event of drawing a bolt produced by the Machines A, B and C respectively. Then
P(E₁) = 25/100, P(E₂) = 35/100, P(E₃) = 40/100, P(E/E₁) = 5/100, P(E/E₂) = 4/100, P(E/E₃) = 2/100.
Therefore, P(E₂/E) = [P(E₂).P(E/E₂)]/[P(E₁).P(E/E₁) + P(E₂).P(E/E₂) + P(E₃).P(E/E₃)]
= [(35/100)(4/100)]/[(25/100)(5/100) + (35/100)(4/100)
+ (40/100)(2/100)]
= [35×4]/[25×5 + 35×4 + 40×2]
= 140/(125 + 140 + 80)
= 140/345
= 28/69.[Ans.]

Q.2. In a bulb factory, machines A, B and C manufacture 60%, 30% and 10% bulbs respectively. 1%, 2% and 3% of the bulbs produced respectively by A, B and C are found to be defective. A bulb is picked up at random from the total production and found to be defective. Find the probability that this bulb was produced by the machine A.

Solution :

[Do yourself. Ans. = 2/5.]

Paper By Mr. M.P.Keshari
Email Id : [email protected]
Ph No. : 09434150289