# Probability

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Question 1 |

A number p is chosen at random from the numbers [-3, 3]. What is the probability that value of p < 2

3/7 | |

5/7 | |

4/7 | |

1/7 |

**Probability**

**Discuss it**

Question 1 Explanation:

Favorable/Total = -ve numbers/total number

Question 2 |

A box contains 6 black, 5 brown and 2 yellow balls. If 2 balls are selected at random, what is the probability that both are black?

4/23 | |

5/26 | |

7/26 | |

8/15 |

**Probability**

**Discuss it**

Question 3 |

4 dices are thrown simultaneously. What is the probability that all the dices show the same number.

1/216 | |

1/36 | |

4/216 | |

4/36 |

**Probability**

**Discuss it**

Question 4 |

An urn contains 12 red and 18 white marbles. Two marbles are drawn without replacement one after another. What is the probability that first is red and second is white?

16/145 | |

9/32 | |

36/145 | |

5/32 |

**Probability**

**Discuss it**

Question 4 Explanation:

Two marbles are drawn without replacement one after another, then the probability that first is red and second is white,

= (12/30)*(18/29) = (2/5)*(18/29) = (2*18) / (5*29) = 36 / 145So, option (C) is correct.

Question 5 |

What is the probability that a number selected from numbers [1,30] is prime number?

1/3 | |

2/7 | |

5/9 | |

5/30 |

**Probability**

**Discuss it**

Question 6 |

A bag contains 21 balls numbered 1 to 21. two balls are drawn one after another without replacement. Find the probability that both balls are even numbered.

3/14 | |

2/3 | |

6/9 | |

1/8 |

**Probability**

**Discuss it**

Question 7 |

A 5 digit number is formed by using digits 1, 2, 3, 4 and 5 with no two digits same. What is the probability that the formed number is divisible by 4?

1/5 | |

4/5 | |

3/5 | |

1/4 |

**Probability**

**Discuss it**

Question 7 Explanation:

To be divisible by 4, the number has to end with the last two digits forming a number divisible by 4. Since there are only 5 digits: 1, 2, 3, 4, 5.
So, possible last two digits according to divisibility by 4:
12
24
32
52
For each of them we now have to put the remaining numbers in front: 3x2x1, therefore 6 combinations for each number set = 6x4 (we have 4 sets from above) = 24.
So, required probability is:
P = Number of desired outcomes / number of possible outcomes
= 24 /(5*4*3*2*1)
= 24 / (5!)
= 24 / 120
= 1/5
So, option (A) is correct.

Question 8 |

A box having 5 black and 3 brown flags. Another box having 4 black and 6 brown flags. If one flag is drawn from each box.
Find the probability that both flags are of different color.

21/40 | |

10/19 | |

3/8 | |

3/10 |

**Probability**

**Discuss it**

Question 9 |

Ram speaks truth in 3/4 of cases and Shyam in 4/5 of cases. In what percent of cases while narrating the same event are they likely to contradict each other?

35% | |

25% | |

20% | |

40% |

**Probability**

**Discuss it**

Question 10 |

What is the probability of getting a sum of 9 from two throws of a dice?

1/3 | |

1/6 | |

1/9 | |

1/12 |

**Probability**

**Discuss it**

Question 10 Explanation:

Total number of outcomes = 6 * 6 = 36.
Total number of favorable outcomes = (3, 6), (4, 5), (5, 4) (6, 3) = 4.
Therefore, required probability = 4/36 = 1/9.

There are 19 questions to complete.