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Probability Questions Answers

Probability Questions Answers (MCQ) listings with explanations includes questions of determining chances of happening an event in different type of problems using different methods which are important for many competitive exams.

1) Bag contain 10 back and 20 white balls, One ball is drawn at random. What is the probability that ball is wh

  • 1) 1
  • 2) 2/3
  • 3) 1/3
  • 4) 4/3
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Ans.   B
Explanation :
Total cases = 10 + 20 = 30 Favourable cases = 20 So probability = 20/30 = 2/3

2) In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor gree

  • 1) 2/3
  • 2) 8/21
  • 3) 3/7
  • 4) 9/22
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Ans.   B
Explanation :
Total number of balls = (8 + 7 + 6) = 21 Let E = event that the ball drawn is neither blue nor green =e vent that the ball drawn is red. Therefore, n(E) = 8. P(E) = 8/21.

3) A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart

  • 1) 1/13
  • 2) 2/13
  • 3) 1/26
  • 4) 1/52
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Ans.   C
Explanation :
Total number of cases = 52 Favourable cases = 2 Probability = 2/56 = 1/26

4) From a pack of 52 cards, 1 card is drawn at random. Find the probability of a face card dra

  • 1) 4/13
  • 2) 1/52
  • 3) 1/4
  • 4) None of above
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Ans.   A
Explanation :
Total number of cases = 52 Total face cards = 16 [favourable cases] So probability = 16/52 = 4/13

5) A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective

  • 1) \begin{aligned} \frac{7}{19} \end{aligned}
  • 2) \begin{aligned} \frac{6}{19} \end{aligned}
  • 3) \begin{aligned} \frac{5}{19} \end{aligned}
  • 4) \begin{aligned} \frac{4}{19} \end{aligned}
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Ans.   A
Explanation :
Please remember that Maximum portability is 1. So we can get total probability of non defective bulbs and subtract it form 1 to get total probability of defective bulbs. So here we go, Total cases of non defective bulbs \begin{aligned} ^{16}C_2 = \frac{16*15}{2*1} = 120 \\text{total cases = } \ ^{20}C_2 = \frac{20*19}{2*1} = 190 \\text{probability = } \frac{120}{190} = \frac{12}{19} \\text{P of at least one defective = } 1- \frac{12}{19} \=\frac{7}{19} \end{aligned}

6) A speaks truth in 75% of cases and B in 80% of cases. In what percentage of cases are they likely to contradict each other, narrating the same incid

  • 1) 30%
  • 2) 35%
  • 3) 40%
  • 4) 45%
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Ans.   B
Explanation :
Let A = Event that A speaks the truth B = Event that B speaks the truth Then P(A) = 75/100 = 3/4 P(B) = 80/100 = 4/5 P(A-lie) = 1-3/4 = 1/4 P(B-lie) = 1-4/5 = 1/5 Now A and B contradict each other = [A lies and B true] or [B true and B lies] = P(A).P(B-lie) + P(A-lie).P(B) [Please note that we are adding at the place of OR] = (3/5*1/5) + (1/4*4/5) = 7/20 = (7/20 * 100) % = 35%

7) From a pack of 52 cards, two cards are drawn together, what is the probability that both the cards are ki

  • 1) 2/121
  • 2) 2/221
  • 3) 1/221
  • 4) 1/13
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Ans.   C
Explanation :
\begin{aligned} \text{Total cases =} ^{52}C_2 \ = \frac{52*51}{2*1} = 1326 \\text{Total King cases =} ^{4}C_2 \\ = \frac{4*3}{2*1} = 6 \ \text{Probability =} = \frac{6}{1326}\ = \frac{1}{221} \\ \end{aligned}

8) A box contains 5 green, 4 yellow and 3 white balls. Three balls are drawn at random. What is the probability that they are not of same colo

  • 1) 52/55
  • 2) 3/55
  • 3) 41/44
  • 4) 3/44
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Ans.   C
Explanation :
\begin{aligned} \text{Total cases =} ^{12}C_3 \\ = \frac{12*11*10}{3*2*1} = 220 \\ \text{Total cases of drawing same colour =} \ ^{5}C_3 + ^{4}C_3 + ^{3}C_3 \\ \frac{5*4}{2*1} + 4 + 1 = 15 \ \text{Probability of same colur =} = \frac{15}{220}\\ = \frac{3}{44} \\ \text{Probability of not same colur =} \ 1-\frac{3}{44}\\ = \frac{41}{44} \end{aligned}

9) In a throw of coin what is the probability of getting hea

  • 1) 1
  • 2) 2
  • 3) 1/2
  • 4) 0
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Ans.   C
Explanation :
Total cases = [H,T] - 2 Favourable cases = [H] -1 So probability of getting head = 1/2

10) In a throw of coin what is the probability of getting tail

  • 1) 1
  • 2) 2
  • 3) 1/2
  • 4) 0
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Ans.   C
Explanation :
Total cases = [H,T] - 2 Favourable cases = [T] -1 So probability of getting tails = 1/2
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