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

## Quiz- Probability problem solving

Start

Congratulations - you have completed

*Quiz- Probability problem solving*. You scored %%SCORE%% out of %%TOTAL%%. Your performance has been rated as %%RATING%%
Your answers are highlighted below.

Question 1 |

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

A | 1/8 |

B | 1/9 |

C | 3 |

D | 8/9 |

Question 1 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 2 |

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

A | 13/14 |

B | 19/23 |

C | 25/54 |

D | 35/45 |

Question 2 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 3 |

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 | 3 |

B | 10 |

C | 5 |

D | 8 |

Question 3 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

Question 4 |

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 | 4/5 |

C | 3/5 |

D | 1/2 |

Question 4 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 5 |

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 | 19/20 |

B | 1/20 |

C | 4/5 |

D | 3/4 |

Question 5 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 6 |

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

A | 17/100 |

B | 19/100 |

C | 1/5 |

D | 1/10 |

Question 6 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 7 |

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 | 3/4 |

D | 1/4 |

Question 7 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 8 |

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

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

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

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

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

Question 8 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 9 |

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 | 1/20 |

B | 1/25 |

C | 4/35 |

D | 4/25 |

Question 9 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 10 |

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 | 1/5 |

C | 1/4 |

D | 1/3 |

Question 10 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 11 |

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 | 1/121 |

B | 1/222 |

C | 1/221 |

D | 1/131 |

Question 11 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 12 |

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

A | 4/13 |

B | 5/13 |

C | 2/13 |

D | 1/13 |

Question 12 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 13 |

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 | 3 |

B | 2/9 |

C | 4/9 |

D | 7/9 |

Question 13 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 14 |

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 | 5/36 |

C | 1/8 |

D | 3/36 |

Question 14 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

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
Get Results

There are 14 questions to complete.

← |
List |
→ |

Return

Shaded items are complete.

1 | 2 | 3 | 4 | 5 |

6 | 7 | 8 | 9 | 10 |

11 | 12 | 13 | 14 | End |

Return

You have completed

questions

question

Your score is

Correct

Wrong

Partial-Credit

You have not finished your quiz. If you leave this page, your progress will be lost.

Correct Answer

You Selected

Not Attempted

Final Score on Quiz

Attempted Questions Correct

Attempted Questions Wrong

Questions Not Attempted

Total Questions on Quiz

Question Details

Results

Date

Score

Hint

Time allowed

minutes

seconds

Time used

Answer Choice(s) Selected

Question Text

All done

Need more practice!

Keep trying!

Not bad!

Good work!

Perfect!

If you liked my contribution, please donate or click on any of advt. a little that will be generated will all be donated for noble cause. [paypal-donation]

>>>> You may be interested to read the following;