Hello Aspirants.

Welcome to Online** Quantitative Aptitude Section** in AffairsCloud.com. Here we are creating question sample in **Probability**, which is common for all competitive exams. We have included Some questions that are repeatedly asked in bank exams !!

**Bag A contains 7 Red Balls, ‘X’ Green Balls, and 5 Yellow Balls. The probability to pick Green Ball at random is 2/5. Another Bag B contains ‘X-3’ Red Balls, ‘X-4’ Yellow Balls and 6 Green Balls. If two balls are picked one after the other from Bag B at random then what is the probability for the Balls to be Red?**

1. 1/21

2. 2/21

3. 3/21

4. 4/21

5. Cannot be determined

###### Answer & Explanation

Answer –**2. 2/21**

**Explanation :**

X/(7+X+5) =2/5

X =8

Bag B = 5 R, 4 Y , 6 G

Probability = 5/15*4/14 = 2/21

**Six individual sock are present in a drawer- Two Red, Two Black and Two White. Pradeep picked one sock randomly to wear. Now he draws another sock from the drawer then what is the probability he draws a sock of same color?**

1. 1/5

2. 1/6

3. 1/30

4. 11/30

5. None

###### Answer & Explanation

Answer –**1. 1/5**

**Explanation :**

In the first draw, he picks anyone sock. Now he is left with 5 socks he has to pick one from them of the same color so 1/5**Born babies to be boy is having 0.52 probability, and to be girl is 0.48. If you have two children, then what is the probability that they are both girls?**

1. 0.2304

2. 0.2704

3. 0.2496

4. 0.7696

5. None

###### Answer & Explanation

Answer –**1. 0.2304**

**Explanation :**

0.48*0.48 = 0.2304

**Anil placed a deck of 52 cards. If you pick two cards what is the probability that both are aces?**

1. 1/169

2. 1/221

3. 1/338

4. 4/663

5. None

###### Answer & Explanation

Answer –**2. 1/221**

**Explanation :**

4/52*3/51**Continuation of question of 5: if Anil fails to draw aces in two chances. Now he is given ten more chances to pick aces, again he failed. Then his friend Ramesh picks two cards from remaining cards, then what is the probability that he exactly draws one ace?**

1. 0.075

2. 0.0923

3. 0.15

4. 0.1846

5. None

###### Answer & Explanation

Answer –**4. 0.1846**

**Explanation :**

4/40*36/39 + 36/40*4/39 = 0.1846**You have drawn a card from 52 cards deck. without replacing it you have drawn another card. Then what is the probability of that card to be King of Diamond?**

1. 1/51

2. 1/52

3. 2/51

4. 3/52

5. None

###### Answer & Explanation

Answer –**1. 1/51**

**Explanation :**

1/52*0 + 51/52*1/51 = 1/51**Suppose children like three types of chocolates Perk, Munch, and 5Star. If they are asked to choose to pick chocolate they have their own preference, one-sixth of children population preference is Perk>Munch>5star. One-sixth of children population preference is Munch>5star>Perk. Similarly remaining four-sixths children preferences follows as per above combinations. If you met a random child and give him chance to pick a chocolate between Munch and Perk. He picked Munch. Now you offer Munch and 5star, what is the probability that he chooses again Munch?**

1. 1/6

2. 1/2

3. 2/3

4. 3/4

5. None

###### Answer & Explanation

Answer –**3. 2/3**

**Explanation :**

Possible preferences

P>M>5

M>5>P

P>5>M

M>P>5

5>P>M

5>M>P

Child picks Munch over Perk P(M5P), P(MP5) or P(5MP)

Then child prefers Munch over 5star P(M5P) P(MP5)

Probability = (1/6+1/6)/(1/6+1/6+1/6) =2/3

**A bag contains 100 tickets, numbered from 1 to 100. If three tickets are picked at random and with replacement, what is the probability that sum of three numbers on the tickets will even number?**

1. 1/2

2. 3/4

3. 1/8

4. 3/8

5. None

###### Answer & Explanation

Answer –**1. 1/2**

**Explanation :**

P(odd) = P(even) = 1/2

Sum of three numbers to be even cases

Even+ even+ even = 1/2*1/2*1/2

Even+ odd+ odd = 1/2*1/2*1/2

Odd+ odd + even = 1/2*1/2*1/2

Odd+ even + odd = 1/2*1/2*1/2

Probability = 1/8+1/8+1/8+1/8 = ½

**Tickets numbered from 1 to 40 are in a bag and one ticket is drawn at random. What is the probability that ticket drawn is a multiple of 3 or 7?**

1. 15/30

2. 16/30

3. 17/30

4. 18/30

5. None

###### Answer & Explanation

Answer –**3. 17/30**

**Explanation :**

Multiples of 3: (3,6,9,12,15,18,21,24,27,30,33,36,39) = 13

Multiples of 7: (7, 14, 21, 28, 35) = 5

Probability = 13+5-1/30 = 17/30

**A Bag contains some White and Black Balls. The probability of picking two white balls one after other without replacement from that bag is 14/33. Then what will be the probability of picking two Black balls from that Bag if bag can hold maximum 15 balls only?**

1. 11/32

2. 14/33

3. 7/33

4. 1/11

5. Cannot be determined###### Answer & Explanation

Answer –**4. 1/11**

**Explanation :**

Wc2/ (B+W)c2 = 14/33

W (W-1)/(W+B)*(B+W-1) = 14/33

Now expressing 14/33 in the above format by multiplying 4 in numerator and denominator

W (W-1)/(W+B)*(B+W-1) = 8*7/12*11 (note = balls <15)

W =8

W+B =12 B =4

Probability = 4c2/12c2 = 1/11

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