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**Solution :-**

*Q11. Option B*

*P(B|A) = P(A n B)/P(A)
P(A n B) = P(B|A) × P(A)
P(A n B) = 0.2 × 0.18
P(A n B) = 0.36*

*Q12. Option A*

*P(A ∩ B) = P(A) + P(B) – P(A **∪** B)
= 8/13 = 5/13 + 7/13 – P(A *

*∪*

*B)*

= P(A

= P(A

*∪*

*B) = 5/13 + 7/13 – 8/13*

= P(A

= P(A

*∪*

*B) = 4/13*

*Q13. Option C*

*P(A|B) = P(A **∪** B)/P(B)
P(A|B) = (6/15)/(4/15) = 3/2.*

*Q14. Option D*

*P(B|A) = P(A **∪** B)/P(A)
P(B|A) = (4/17)/(6/17) = 4/6 = 2/3.*

*Q15. Option D*

*Let E and F denote respectively the events that first and second ball drawn are black. We have to find P(E n F) or P (EF).
Now P(E) = P (black ball in first draw) = 10/15
Also given that the first ball drawn is black, i.e., event E has occurred, now there are 9 black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability of F given that E has occurred.
That is P(F|E) = 9/14
By multiplication rule of probability, we have
P (E n F) = P(E) P(F|E)
= 10/15 × 9/14 = 3/7*

*Q16. Option C*

*Let Q denote the event that the card drawn is queen and A be the event that
the card drawn is an ace. Clearly, we have to find P (QQA)
Now P(Q) = 4/52
Also, P (Q|Q) is the probability of second queen with the condition that one queen has
already been drawn. Now there are three queen in (52 – 1) = 51 cards.
Therefore P(Q|Q) = 3/51
P(A|QQ) is the probability of third drawn card to be an ace, with the condition
that two queens have already been drawn. Now there are four aces in left 50 cards.
Therefore P(A|QQ) = 4/50
By multiplication law of probability, we have
P(QQA) = P(Q) P(Q|Q) P(A|QQ)
= 4/52 × 3/51 × 4/50
= 2/5525.*

*Q17 . Option B*

*We know that the sample space is S = {1, 2, 3, 4, 5, 6}
Now G = { 3, 6}, F = { 2, 4, 6} and E n F = {6}
Then P(G) = 2/6 = 1/3
P(H) = 3/6 = 1/2 and P(G ∩ H) = 1/6
P(G n H) = P(G). P (H)
G and H are independent events.*

*Q18. Option C*

*If all the 36 elementary events of the experiment are considered to be equally likely, we have
P(A) = 18/36 = 1/2
= and P(B) = 18/36 = 1/2
Also P(A n B) = P (odd number on both throws)
= 9/36 = 1/4
Now P(A) P(B) = 1/2 × 1/2 = 1/4
P(A n B) = P(A) × P(B)
A and B are independent events*

*Q19 Option C*

*2 ^{6} = 64, ( 1 = 2, 2 = 4, 4 = 16, 16 = 32, 32=64 times )
*

_{2}*= 6!/2! = (6 × 5 × 4 × 3 × 2 × 1)/ (6 – 2)! × (2 × 1) = 15*

Probability = 15/64.

Probability = 15/64.

*Q20. Option B*

*P (A n B) = P(A) . P(B)
P (A n B) = 4/5 . 2/5
P (A n B) = 8/25.*

* *