1.

What is the probability that when a hand of 5 cards is drawn from a well shuffled deck of 52 cards,

**Answer: Option 'D'**

^{n}C_{r} = n!/(n-r)!r!

Total number of possible hands = ^{52}C^{5}

^{52}C^{5} = 2274090

Number of hands with 4 Queens = ^{4}C_{4} × ^{48}C^{1}

^{4}C_{4} = 24

^{4}C_{1} = 48

(other 1 card must be chosen from the rest 48 cards)

Hence P (a hand will have 4 Queens) = (^{4}C_{4} × ^{48}C^{1})/^{52}C^{5} = 192/379015.

2.

Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are queens and the third card drawn is an ace?

**Answer: 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.**

3.

If P(A) = 6/17, P(B) = 5/17, and P(A ∪ B) = 4/17 Find P(B|A)?

**Answer: Option 'D'**

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

4.

6 Coins are tossed simultaneously. find the probability to get 2 hands

**Answer: Option 'C'**

**2 ^{6} = 64, ( 1 = 2, 2 = 4, 4 = 16, 16 = 32, 32=64 times )
6_{c}_{2} = 6!/2! = (6 × 5 × 4 × 3 × 2 × 1)/ (6 - 2)! × (2 × 1) = 15
Probability = 15/64.**

5.

A die is thrown. If G is the event 'the number appearing is a multiple of 3' and H be the event 'the number appearing is even' then find whether G and H are independent ?

**Answer: 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.**

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