This a quiz based on the concept discuss in the earlier post : Concept of Probability

## Quiz- Probability problem solving

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Question 1 |

A box contains 5 pairs of shoes, If two shoes are selected at random, what it is the probability that they will form a pair, assume all the pairs are different?

A | 2/9 |

B | 4/9 |

C | 7/9 |

D | 3 |

Question 1 Explanation:

since there are 5 pairs, the no. of ways to select a pair is 5C2 as per comination formula= 5 x 4 =20, no. of ways to select any 2 shoes is 10C2 = 10 x 9 =90, hence proability to select a pair of shoe in random selection of 2 shoes is = 20/90 = 2/9

Question 2 |

What is the probability that the total of two dice will be greater than or equal to 8

**given that**the first die is a 4?A | 1/3 |

B | 1/2 |

C | 4/5 |

D | 3/5 |

Question 2 Explanation:

This is a case of conditional probability. The possible outcome with 4 in the first die are (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6). The desired outcome are 3 sets which are greater than or equal to 8. So the probability is 3/6 = 1/2

Question 3 |

What is the probability that a card from a deck will be either an ace or a king?

A | 4/13 |

B | 1/13 |

C | 2/13 |

D | 5/13 |

Question 3 Explanation:

Since question has OR, the proabaility events are mutually exclusive. The probability will add up, the probability to draw Ace = 4/52 and King = 4/52
Adding both = 8/52

Question 4 |

What is probability of only 2 dices showing same numbers when the 5 dies are thrown at a time?

A | 13/14 |

B | 35/45 |

C | 25/54 |

D | 19/23 |

Question 4 Explanation:

The total number of possible outcome in throw = 6 ^ 5. The combinations for favorable outcome is as follows: 2 dice with same face can be selected in 5C2 = 10 ways, these can take any of 6 face values, rest 3 the will have option of 5, 4 and 3, so favorable outcome can be 10x 6 x 5 x 4 x 3
Probability = (10 x 6 x 5 x 4 x 3)/ 6 x 6 x 6 x 6 x 6 = 25/54

Question 5 |

A book contain 100 Pages, what is probability that a page no. with digit 2 will be opened in first attempt?

A | 19/100 |

B | 17/100 |

C | 1/10 |

D | 1/5 |

Question 5 Explanation:

Between 1 and 100 there are 19 numbers that 2 in any form, ex. 2, 21, 22, 23, 32, 42 etc. so the probability is 19/100

Question 6 |

There are 10 employees in the office what is probability that atleast 2 employees with share thier brithday in a year?

A | 1- 365!/ (355! x 365^10) |

B | 1- (10!/365!) |

C | 1- (10!/(365^10)) |

D | (10/365)* (355/364) |

Question 6 Explanation:

Since questions ask atleast it easier to find the proabability of none of employees sharing birthday and subtract with 1. The first one can have birthday on any day so it is 365/365, but second one can have on 364 days so it is 364/365 and so on so we probability = (365/365) x (364/365) x (363/365)... up to 10 time. = 365!/ (355! x 365^10) , the probability is equal to 1- 365!/ (355! x 365^10)

Question 7 |

What is the probability that when a pair of fair six-sided dice are thrown, the sum of the number equals 12 ?

A | 1/36 |

B | 3/36 |

C | 1/8 |

D | 5/36 |

Question 7 Explanation:

total number of possible outcomes= 36, outcome can be 12 only when both the dice throws up 6 and 6, hence there is only 1 possible outcome

Question 8 |

He is a big liar and speaks truth only once in four time. If after throwing coin he says it's a head what is probability it is actually a head?

A | 1/2 |

B | 5/4 |

C | 1/4 |

D | 3/4 |

Question 8 Explanation:

Probability he is telling truth = 1/4, Probability it is head = 1/2, So probability of fav. outcome = (1/2)x (1/4) = 1/8. The total no. of outcome either it is head and he telling truth or it is not a head and he is telling lie = (1/2) x (1/4) + (1/2) x(3/4) = 1/2
So the proability is (1/8)/ (1/2) = 1/4

Question 9 |

What is the probability that the the outcome has a sum of 5 when a pair of six sided dice is thrown?

A | 3 |

B | 1/8 |

C | 1/9 |

D | 8/9 |

Question 9 Explanation:

The total no. of outcomes = 36, the possible combinations are (1,4), (4,1), (2,3) , (3,2) hence the probability is 4/36

Question 10 |

A combination lock has 3 rings with numbers 0-9 inscribed, If Sudha has 5 minute in which she can try for 50 combinations, what is her probability of failure?

A | 3/4 |

B | 4/5 |

C | 19/20 |

D | 1/20 |

Question 10 Explanation:

No. of possible combination in lock = 10 x 10 x 10 = 1000. There is only correct combination, that is there are 999 wrong combination, probability of failure in first attempt = 999/1000 2nd 998/999 and so on for 50 time. all the denominator except first one and numerator except last one will get cancelled. so answer will 950/1000 = 19/20

Question 11 |

What is the probability that the position in Vowel will appear in the same positions when the letters of the word MATHS are re-arranged?

A | 1/6 |

B | 1/4 |

C | 1/5 |

D | 1/3 |

Question 11 Explanation:

Total number of possible arrnagements = 5! = 5x4x3x2x1 = 120, if position of only vowel A is fixed to 2nd place, then other 4 can be arranged in 4! way. 4x3x2x1 = 24, so the probability is = 24/120 = 1/5

Question 12 |

If there are 20 Black and White marbles in a jar, and the ratio of Black to White marbles is 2:3, what is the probability that, drawing twice, one will select two black marbles if he or she return the marbles after each draw?

A | 1/25 |

B | 4/35 |

C | 4/25 |

D | 1/20 |

Question 12 Explanation:

If the black marbles are 2x then white marbles are 3x, in total there are 20, so x = 4, black marbles = 8, white marbles = 12. Probability for to draw the black marble is 8/20 in first draw and since the marble is returned back, probability to draw 2nd is 8/20. So total probaility = (2/5) x (2/5) = 4/25

Question 13 |

If one draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be King?

A | 1/221 |

B | 1/121 |

C | 1/222 |

D | 1/131 |

Question 13 Explanation:

Probability to draw a king in first attempt = 4/52 as there are 4 Kings in deck of 52 cards. Probability to draw a 2nd King without replacing card is 3/51 as there are now 3 kings and 51 cards. (4/52) x (3/51) = 1/221

Question 14 |

In a class room, how many least numbers of pupil are requied to have the probability of greater than 0.5 of atleast 1 child in group to born in leap year?

A | 8 |

B | 10 |

C | 3 |

D | 5 |

Question 14 Explanation:

Proability of randomly selecting a pupil NOT born in leap year = 3/4. If we take 2 pupil, the probability they are not born in leap year = (3/4) x (3/4) = 9/16. The probability that atleast one of the two is born in leap year = 1- (9/16) = 7/16 = 0.4... which is small er than 0.5. Lets try with 3 pupils, thus we have 1- (27/64) = 37/64. which is greater than 0.5 hence the answer is 3

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