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# Probability Questions And Answers For CAT Set-4 PDF:

Download Probability Questions And Answers For CAT Set-4 PDF. Practice important Probability Questions with detailed answers and explanations. These questions are based on previous CAT question papers.

Question 1:There are three identical bags. In the first bag, there are 6 black balls, 3 blue balls and 1 golden ball. In the second bag there are 3 black balls, 3 blue balls and 2 golden balls. In the third bag, there are 2 black balls and 3 blue balls. If Raju picks up a ball from one of the bags, find the probability that it is a golden ball.

a) $\frac{1}{6}$

b) $\frac{7}{60}$

c) $\frac{2}{15}$

d) $\frac{3}{20}$

Question 2: The number of ways of reaching origin from point (3,3) such that a person can take only one step at a time either along a x axis or along a y axis is

a) 28
b) 20
c) 45
d) 55

Question 3: There are three winners from the race with 1st, 2nd and 3rd position respectively. There are 8 prizes of different values with exactly one prize given to each winner. Find the number of ways of distributing 3 prizes such that the value of the prize of the first person is more than the that of the second person and the value of the prize of the second person is more than that of the third person?

a) 56
b) 51
c) 21
d) 76

Question 4: There are five prize winners – A, B, C, D and E – and 7 prizes – First, second, third, fourth, fifth, sixth and seventh. In how many ways can these prizes be distributed such that all the prizes that A get are better than those that B gets, all the prizes that B get are better than those that C gets and so on (Everyone gets at least 1 prize?

a) 10
b) 15
c) 20
d) 25

Question 5: What is the rank of the word FORMAT in the dictionary order among all the 6-letter words that can be formed using the letters in the given word?

a) 183
b) 184
c) 185
d) 186

Probability of picking up each bag is $\frac{1}{3}$
Probability of picking a golden ball in first bag = $\frac{1}{10}$
Probability of picking a golden ball in second bag = $\frac{1}{4}$
Probability of picking a golden ball in third bag = 0
Total probability = ($\frac{1}{3}$*$\frac{1}{10}$) + ($\frac{1}{3}$*$\frac{1}{4}$) + ($\frac{1}{3}$*0)
= $\frac{1}{30}$ + $\frac{1}{12}$
= $\frac{7}{60}$

There are 6 steps in total out of whih 3 are horizontal and 3 are vertical:HHHVVV.
Number of ways of reaching origin is = 6!/(3!*3!) = 20

Let the prizes be a,b,c,d,e,f,g,h with a>b>c>d>e>f>g>h.
Now if we select any 3 prize out of 8, these 3 prizes can be arranged in only 1 way with 1st position getting the highest prize, 2nd position getting the second prize and the third position getting the lowest valued prize.
Number of ways =$^8C_3$ = 56

1 _ 2 _ 3 _ 4 _ 5 _ 6 _ 7
We need to put 4 persons in these 6 places such that all the prizes that come between a person X and the person before him belong to person X.
There are 6 gaps and 4 persons to be put. The remaining prizes to the right of the fourth person belong to the fifth person.
This can be done in $^6C_4$ = 15 ways.