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Probability (Hard Level) - CAT MCQ


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10 Questions MCQ Test Quantitative Aptitude (Quant) - Probability (Hard Level)

Probability (Hard Level) for CAT 2024 is part of Quantitative Aptitude (Quant) preparation. The Probability (Hard Level) questions and answers have been prepared according to the CAT exam syllabus.The Probability (Hard Level) MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Probability (Hard Level) below.
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Probability (Hard Level) - Question 1

Out of a pack of 52 cards one is lost; from the re- mainder of the pack, two cards are drawn and are found to be spades. Find the chance that the missing card is a spade.

Detailed Solution for Probability (Hard Level) - Question 1

This problem has to be treated as if we are selecting the third card out of the 50 remaining cards. 11 of these are spades. Hence, 11/50.

Probability (Hard Level) - Question 2

A pair of fair dice are rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is,

Detailed Solution for Probability (Hard Level) - Question 2

We do not have to consider any sum other than 5 or 7 occurring.  A sum of 5 can be obtained by any of [4 + 1, 3 + 2, 2 + 3, 1 + 4] 
Similarly a sum of 7 can be obtained by any of [6 + 1, 5 + 2, 4 + 3, 3 + 4, 2 + 5, 1 + 6]
n(E) = 4, n(S) = 6 + 4
P = 0.4

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Probability (Hard Level) - Question 3

In a bag, there are ten black, eight white and five red balls. Three balls are chosen at random and one is found to be black. The probability that the rest two are white is. Find the probability that the remaining two balls are white.

Detailed Solution for Probability (Hard Level) - Question 3

The required probability would be given by the event 
definition:First is white and second is white
= 8/22 X 7/21 = 4/33.

Probability (Hard Level) - Question 4

A bag contains 15 tickets numbered 1 to 15.A Ticket is drawn and replaced. Then one more ticket is drawn and replaced. The probability that first number drawn is even and second is odd is

Detailed Solution for Probability (Hard Level) - Question 4

In the first draw, we have 7 even tickets out of 15 and in the second we have 8 odd tickets out of 15.
Thus, (7/15) X (8/15) = 56/225

Probability (Hard Level) - Question 5

Three faces of a dice are yellow, two faces are red and one face is blue. The dice is tossed three times.Find the probability that the colours yellow, red and blue appear in the first, second and the third toss respectively:

Detailed Solution for Probability (Hard Level) - Question 5

Total we have 6 faces of dice,
Yellow - 3 faces;  Probability of this face is  3/6
Red -  2 faces; Probability of this face is  2/6
Blue - 1 face; Probability of this face is  1/6
Yellow and Red and Blue = (3/6) X (2/6) X (1/6) = (6/216) = 1/36.

Probability (Hard Level) - Question 6

There are 5 envelopes corresponding to 5 letters. If the letters are placed in the envelopes at random, what is the probability that all the letters are not placed in the right envelopes?

Detailed Solution for Probability (Hard Level) - Question 6

All four are not in the correct envelopes means that at least one of them is in the wrong envelope. A little consideration will show that one letter being placed in a wrong envelope is not possible, since it will have to be interchanged with some other letter. Since, there is only one way to put all the letters in the correct envelopes, we can say that the event of not all four letters going into the correct envelopes will be given by
5!–1=119

Probability (Hard Level) - Question 7

A fair coin is tossed 10 times. Find the probability that two Heads do not occur consecutively. 

Detailed Solution for Probability (Hard Level) - Question 7

We can have a maximum of 5 heads. 
For 0 head, P(E)=(1/210)X1 
For 1heads, P(E)=(1/210)X1
 For 2 heads and for them not to occur consecutively we will need to see the possible distribution of 8 tails and 2 heads.
Since the 2 heads do not need to occur consecutively, this would be given by (All – 2heads together)
 = (10C8 – 9)
P(E)=(10C8-9) / 210 = 1/23

Probability (Hard Level) - Question 8

The probability that a contractor will get a plumbing contract is 2/3 and the probability that he will get an electric contract is 5/9. If the probability of getting at least one contract is 4/5, what is the probability that he will get both the contracts?

Detailed Solution for Probability (Hard Level) - Question 8


From the venn diagram we get:
Using the formula P(A) + P(B) = P (A U B) + P (A ∩ B)
Here we have P (A ) = ⅔
P(B) = 5/9
P(A U B) = ⅘, we need to find P (A ∩ B)
(2/3) + (5/9) – x = 4/5 
x = (2/3) + (5/9) – (4/5) = 19/45

Probability (Hard Level) - Question 9

Two small squares on a chess board are chosen at random. Find the probability that they have a com- mon side:

Detailed Solution for Probability (Hard Level) - Question 9

The common side could be horizontal or vertical. Accordingly, the number of ways the event can occur is.
n(E)=8X7+8X7=112
n (S) = 64C2
= 2X8X7X2/64 X 63 = 1 / 18 

Probability (Hard Level) - Question 10

In a room there are 7 people. The chance that two of them were born on the same day of the week is

Detailed Solution for Probability (Hard Level) - Question 10

This can be got by taking the number of ways in which exactly two people are born on the same day divided by the total number of ways in which 7 people can be born in 7 days of a week. For the first part select two people from 7 in 7C2 ways & select a day from the week on which they have to be born in 7C1 ways & for the remaining 5 people select 5 days out of the remaining six days of the week & then the number of arrangements of these 5 people in 5days- thus a total of 7C2 X 7C1 X 6C5 X 5!ways. 
Also, the number of ways in which seven people can  be born on 7 days would be given by 77 . Hence, the answer is given by: (7C2 X 7C1 X 6C5 X 5!/77
= 21 X 7 X 6 X 120/77=3 X 6 X 120/75 =2160/75

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