Probability Questions For CAT PDF Set-3

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Probability Questions For CAT PDF Set-3
Probability Questions For CAT PDF Set-3

Probability Questions For CAT PDF Set-3:

Download Probability questions and answers for CAT PDF Set-3. This is an list of some important must solve problems for CAT exam with solutions.

Download Probability Questions For CAT PDF Set-3

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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) 12
b) 15
c) 10
d) 20

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

Question 6: How many 8-letter words can be formed using the the letters in the word TRUTHFUL?

a) 10070
b) 10090
c) 10080
d) 10060

Question 7: Five different prizes are to be distributed among 3 students A,B,C such that each student gets at least 1 prize. Find the total number of ways in which prizes can be distributed?

a) 150
b) 100
c) 130
d) 120

Question 8: If there are a total of 24 marbles with Ravi, Rathod, Rahul and Ramu, then what is the probability that Ravi has exactly 8 marbles?

a) $\frac{13}{325}$

b) $\frac{17}{325}$

c) $\frac{19}{325}$

d) $\frac{21}{325}$

Question 9: There are 2 identical bags, B1 and B2. There are 6 blue balls and 9 black balls in B1, and 10 blue balls and 8 black balls in B2. If Ramu picks up 2 balls randomly from a bag and both are blue, then what is the probability that both the balls are picked from B1?

a) $\frac{1}{3}$

b) $\frac{16}{51}$

c) $\frac{17}{52}$

d) $\frac{2}{5}$

Question 10: In how many ways can 18 identical plates be divided into 3 groups such that each group must at least have one plate?

a) 25
b) 27
c) 22
d) 30

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1) Answer (B)

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)
= $\frac{1}{30}$ + $\frac{1}{12}$
= $\frac{7}{60}$

2) Answer (B)

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

3) Answer (A)

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

4) Answer (B)

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.

5) Answer (A)

Ascending order of letters => A, F, M, O, R and T.
720 words can be formed such that each of the letters is in the first position for 120 words.
=> The words that start with F will have a rank between 121-240 both inclusive.
Each of these 120 words have 24 words with second letter the same.
O is the second letter => ranking will be between 169-192.
Similarly R is the third letter => Ranking will be between 181-186.
M is the fourth letter => Ranking will be between 183-184.
A is the fifth letter => Rank is 183.

6) Answer (C) 

There are eight letters of which only 6 are unique. The remaining two are repeated.
=> Number of words that can be formed = $\frac{8!}{2!*2!}$ = 10080

7) Answer (A)

Since the prizes are different, the required distribution would take place in (1,1,3), (2,2,1) ways.
Let us consider the first arrangement – (1,1,3)
There are 3 ways in which the person who gets 3 prizes can be selected. The prizes to be distributed to him can be selected in 5C3 ways. Now the remaining 2 prizes can be distributed to the 2 persons in 2! ways.
Number of ways = 3*5C3*2 = 3 *10* 2 = 60 ways
Let us consider the second arrangement – (2,2,1)
There are 3 ways in which we can select the person who gets 1 prize. The prize for him can be selected in 5C1 = 5 ways. The remaining 2 persons get 2 prizes each. So, we have to select only the ways in which 2 prizes can be selected. 2 prizes can be selected from the 4 available prizes in 4C2 = 6 ways.
Number of ways = 3*5C1*6 = 3*5*6 = 90 ways.
Total number of ways = 60 + 90 = 150.

8) Answer (B)

Let the number of marbles with Ravi, Rathod, Rahul and Ramu be a, b, c and d respectively.
a + b + c + d = 24
a = 8
=> b + c + d = 16
=> Number of ways possible = $^{16+3-1}C_{3-1}$ = $^{18}C_2$ = 153
Total number of ways 24 marbles can be diistributed is $$^{24+4-1}C_{4-1}$ = $^{27}C_3$ = 2925
Probability = $\frac{153}{2925}$ = $\frac{17}{325}$

9) Answer (C) 

Identical bags => P(B1) = P(B2) = $\frac{1}{2}$
B1 has 15 balls, of which 6 are Blue. Probability that both the balls that are picked up from B1 are blue = $\frac{^6C_2}{^{15}C_2}$ = $\frac{1}{7}$
=> P($\frac{Both-Blue}{B1}$) = $\frac{1}{7}$
B2 has 18 balls, of which 10 are Blue. Probability that both the balls that are picked up from B2 are blue = $\frac{^{10}C_2}{^{18}C_2}$ = $\frac{5}{17}$
=> P($\frac{Both-Blue}{B2}$) = $\frac{5}{17}$
Using Bayes theorem, we know that:
P($\frac{B1}{Both-Blue}$) = $\frac{P(\frac{Both-Blue}{B1})*P(B1)}{\frac{Both-Blue}{B1})*P(B1) + \frac{Both-Blue}{B2})*P(B2)}$
=> P($\frac{B1}{Both-Blue}$) = $\frac{\frac{1}{7}*\frac{1}{2}}{\frac{1}{7}*\frac{1}{2} + \frac{5}{17}*\frac{1}{2}}$ = $\frac{17}{52}$

10) Answer (B)

Let the number of plates in the three groups be x, y and z.
If x = 1, then the remaining 17 plates can be distributed between y and z is 8 ways => (1,16), (2,15), (3,14), (4,13), (5,12), (6,11), (7,10), (8,9)
If x = 2, then the remaining 16 plates can be distributed between y and z is 7 ways => (2,14), (3,13), (4,12), (5,11), (6,10), (7,9), (8,8)
If x = 3, then the remaining 15 plates can be distributed between y and z is 5 ways => (3,12), (4,11), (5,10), (6,9), (7,8)
If x = 4, then the remaining 17 plates can be distributed between y and z is 4 ways => (4,10), (5,9), (6,8), (7,7)
If x = 5, then the remaining 17 plates can be distributed between y and z is 2 ways => (5,8), (6,7)
If x = 6, then the remaining 17 plates can be distributed between y and z is 1 ways => (6,6)
=> Total number of ways = 8 + 7 + 5 + 4 + 2 + 1 = 27

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