Probability -Aptitude

1.

Let A and B be independent events with P (A) = 0.13 and P(B) = 0.3. Find P(B/A)? 

   A.) 2.34
   B.) 1.3
   C.) 0.3
   D.) None of these

Answer: Option 'C'

P(B/A) = P (A n B)/P(A) 
Here, P (A n B) = P(A) × P(B) = 0.13 × 0.3 = 0.39 
P(B/A) = 0.39/0.13 = 0.3

2.

An unbiased die is thrown twice. Let the event A be ‘odd number on the first throw’ and B the event  ‘odd number on the second throw’. Check the independence of the events A and B. 

   A.) A or B are independent events
   B.) A and B are not independent events
   C.) A and B are independent events
   D.) None of these

Answer: 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

3.

A die is thrown three times. Events X and Y are defined as below:
X : 4 on the third throw 
Y : 6 on the first and 5 on the second throw 
What is the probability of X given that Y has already occurred. 

   A.) 2/6
   B.) 1/6
   C.) 1/5
   D.) None of these

Answer: Option 'B'

The sample space has 216 outcomes.
Now X = (1,1,4) (1,2,4) ... (1,6,4) (2,1,4) (2,2,4) ... (2,6,4)
(3,1,4) (3,2,4) ... (3,6,4) (4,1,4) (4,2,4) ...(4,6,4)
(5,1,4) (5,2,4) ... (5,6,4) (6,1,4) (6,2,4) ...(6,5,4) (6,6,4)
Y = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)}
and X n Y = {(6,5,4)}.
Now P(Y) = 6/216 
and P (X n Y) = 1/216 

4.

If P (A) = 0.18, P (B) = 0.5 and P (B|A) = 0.2, find P(A n B)? 

   A.) 0.32
   B.) 0.36
   C.) 0.16
   D.) 0.64

Answer: 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

5.

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?

   A.) 2/5530
   B.) 3/5525
   C.) 2/5525
   D.) 4/5525

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.

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