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Test Level 2: Probability - 2 - CAT MCQ


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20 Questions MCQ Test - Test Level 2: Probability - 2

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Test Level 2: Probability - 2 - Question 1

If four whole numbers taken at random are multiplied together, find the probability that the last digit in the product is 1, 3, 7, or 9.

Detailed Solution for Test Level 2: Probability - 2 - Question 1

The last digit in the product can be one of the required digits, if and only if, the last digit of each of the numbers is one of the numbers 1, 3, 7 or 9.
Probability of choosing each of the four numbers is 4/10.
As the last digit of required numbers can be chosen in four different ways and total number of ways of choosing a digit at units place is 10,

Test Level 2: Probability - 2 - Question 2

The numbers 1, 2, 3, …., 100 are written on 100 cards with one number on each card. The cards are placed in a hat and one card is selected. The sizes and shapes of the cards are such that the probability of selecting the card labelled with the number 'n' is equal to 'n' times the probability of selecting the card labelled 1. What is the probability that the card labelled 50 is selected?

Detailed Solution for Test Level 2: Probability - 2 - Question 2

Let 'x' be the probability, that card labelled '1' is selected.
Probability that card labelled '2' is selected = 2x
Probability that card labelled '3' is selected = 3x and so on
Now, x + 2x + 3x + … + 100x = 1, which means x = 1/5050

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Test Level 2: Probability - 2 - Question 3

Cards are drawn one by one from a deck of 52 cards without replacement. What is the probability that four cards are drawn before the first ace (fifth card) is drawn?

Detailed Solution for Test Level 2: Probability - 2 - Question 3

Required probability = P (drawing a non-ace each of the first four times and drawing an ace the fifth time) =

Test Level 2: Probability - 2 - Question 4

Find the probability that a leap year selected at random contains either 53 Sundays or 53 Mondays.

Detailed Solution for Test Level 2: Probability - 2 - Question 4

A leap year has 366 days, or 52 weeks + 2 days. This means that 52 Sundays will be there, but for the 53rd, we have to evaluate the possibility of the two extra or odd days to fall on Sunday or Monday.
The possible outcomes for two odd days can be (Mon, Tue); (Tue, Wed); (Wed, Thu); (Thu, Fri); (Fri, Sat); (Sat, Sun) and (Sun, Mon).
Number of favourable cases = 3
Required probability = 3/7

Test Level 2: Probability - 2 - Question 5

A candidate is to be selected for an interview for two posts. The number of candidates for the first post is 5 and that for the second post is 6. Find the probability of getting at least one job.

Detailed Solution for Test Level 2: Probability - 2 - Question 5

Probability that the candidate does not get an offer from the first interview is 4/5.
Probability that the candidate does not get an offer from the second interview is 5/6.
Total probability of not getting an offer 

Hence, probability of getting at least one job 

Test Level 2: Probability - 2 - Question 6

Two cards are drawn in succession from a pack of 52 cards. The first card should be a Queen and the second should be a King. What is the probability of doing so if the first card is replaced?  

Detailed Solution for Test Level 2: Probability - 2 - Question 6

Probability of drawing a Queen = 4/52 = 1/13
Since the first card is replaced, the pack will again have 52 cards. So, the probability of drawing a King = 4/52 = 1/13
Both the events are independent, hence the probability of drawing both cards in succession 

Test Level 2: Probability - 2 - Question 7

Six unbiased coins are tossed simultaneously. Find the probability that at least two heads occur.

Detailed Solution for Test Level 2: Probability - 2 - Question 7

Probability that one head occurs

Probability that all tails occur 

Hence, probability that at least two heads occur 

Test Level 2: Probability - 2 - Question 8

Two numbers a and b are chosen at random from the set of first 30 natural numbers. The probability that a2 - b2 is divisible by 3 is

Detailed Solution for Test Level 2: Probability - 2 - Question 8

The total number of ways of choosing two numbers out of 1, 2, 3 ........, 30 is
30C2 = 435
⇒ Exhaustive number of cases = 435
Since a2 - b2 is divisible by 3 if either a and b both are divisible by 3 or none of a and b is divisible by 3, so the favourable number of cases = 10C2 + 20C2 = 235
Hence, the required probability = 235/435 = 47/87

Test Level 2: Probability - 2 - Question 9

A bag contains 5 square boxes and 7 cylindrical boxes. Two boxes are drawn at random. Find the probability that they are the same shape.  

Detailed Solution for Test Level 2: Probability - 2 - Question 9

S = sample space. n(s) = No. of ways of drawing two boxes out of (5 + 7) boxes = 12C2 = (12 × 11)/(2 × 1) = 66.
E = Event of getting both boxes of the same shape.
n(E) = No. of ways of drawing (2 boxes out of 5 or 2 boxes out of 7) = 5C2 + 7C2 = (5 × 4)/(2 × 1) + (7 × 6)/(2 × 1) = 31.
P(E) = n(E)/n(S) = 31/66.

Test Level 2: Probability - 2 - Question 10

Consider a pack of 52 cards. One card is drawn at random. What is the probability of the card being a heart or a seven?  

Detailed Solution for Test Level 2: Probability - 2 - Question 10

S = Sample space = 52.
There are 13 hearts including 1 seven.
There are 3 more sevens in spade, club and diamond.
E = event of getting a seven or a heart.
n(E) = 13 + 3 = 16.
P(E) = n(E)/n(S) = 16/52 = 4/13

Test Level 2: Probability - 2 - Question 11

Two cards are drawn one by one at random without replacement from a pack of 52 cards. What is the probability that the cards drawn are queens?

Detailed Solution for Test Level 2: Probability - 2 - Question 11

Here, the sample space is 52.
Probability that the first card is a queen = 4/52 = 1/13
Probability that the second card is a queen = 3/51
Hence, probability that the cards drawn are queens 

Test Level 2: Probability - 2 - Question 12

Two fair dice are thrown. What is the probability that the number of dots on the first dice is greater than 3 and that on the second is greater than 4?

Detailed Solution for Test Level 2: Probability - 2 - Question 12

A: The first dice shows 4, 5 or 6 dots.
B: The seconds dice shows 5 or 6 dots.
P(A) = 3/6 = 1/2
P(B) = 2/6 = 1/3
So, the required probability is
 

Test Level 2: Probability - 2 - Question 13

In a room, there are 15 men and 10 women. Three of them are selected at random. The probability that 1 woman and 2 men are selected is

Detailed Solution for Test Level 2: Probability - 2 - Question 13

S = Sample space and E = event of choosing 1 woman and 2 men.
n(S) = Number of ways of selecting 3 people out of 25

n(E) = (10C1 × 15C2) = 1050.
Therefore, P(E) = n(E)/n(S) = 1050/2300 = 21/46

Test Level 2: Probability - 2 - Question 14

In a box, there are 5 red, 9 green and 6 black balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

Detailed Solution for Test Level 2: Probability - 2 - Question 14

Total number of balls = 5 + 9 + 6 = 20.
E = event that the ball drawn is neither red nor green = event that the ball drawn is black = 6.
Therefore, P(E) = 6/20.

Test Level 2: Probability - 2 - Question 15

In a simultaneous throw of a pair of dice, find the probability of getting an even number on one and a multiple of 3 on the other.  

Detailed Solution for Test Level 2: Probability - 2 - Question 15

Elementary events associated to the random experiment of throwing two dice are: 
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4),(4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Total no. of events = 36
Let A be the event of getting an even number on one and a multiple of 3 on the other.
A : (2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)
So, favourable no. of events = 11
Hence, probability = (No. of favourable events)/(Total no. of events)
Probability = 11/36

Test Level 2: Probability - 2 - Question 16

A bag contains 4 yellow balls, 5 green balls and 3 red balls. What is the probability of drawing a red or a yellow ball?

Detailed Solution for Test Level 2: Probability - 2 - Question 16

Total number of balls = 4 + 3 + 5 = 12
Number of balls which are either red or yellow = 3 + 4 = 7
Here, the total sample space is 12 and the event value is 7.
Hence, probability of getting a red or a yellow ball = 7/12

Test Level 2: Probability - 2 - Question 17

There are two dice; red and green. What is the probability that on roll the red die throws up a prime number which is divisible by the number thrown up on the roll of the green die?  

Detailed Solution for Test Level 2: Probability - 2 - Question 17

There are 3 cases possible as follows:
Case 1:
The red die throws up a 2 and the green die throws up a 2 or a 1, the probability of which is (1/6)(2/6) = 2/36 = 1/18

Case 2:
The red die throws up a 3 and the green die throws up a 3 or a 1, the probability of which is (1/6)(2/6) = 2/36 = 1/18

Case 3:
The red die throws up a 5 and the green die throws up a 5 or a 1, the probability of which is (1/6)(2/6) = 2/36 = 1/18
Thus, the required probability = (1/18) + (1/18) + (1/18) = 3 × (1/18) = 1/6

Test Level 2: Probability - 2 - Question 18

Two fair dices are thrown. Given that the sum of the dice is less than or equal to 4, find the probability that only one dice shows two.

Detailed Solution for Test Level 2: Probability - 2 - Question 18

The possible outcomes are:
(1, 1); (1, 2); (2, 1), (2, 2); (3, 1); (1, 3).
Out of six cases, in two cases there is exactly one ‘2’
Thus, the correct answer is 2/6 = 1/3.

Test Level 2: Probability - 2 - Question 19

There are two bags, one of them contains 5 red and 7 white balls and the other 3 red and 12 white balls, and a ball is to be drawn from one or the other of the two bags. Find the chance of drawing a red ball.

Detailed Solution for Test Level 2: Probability - 2 - Question 19

The event can be defined as:
First bag is selected and red ball is drawn.
or second bag is selected and red ball is drawn.
1/2 X 5/12 + / X 3/15 = (5/24) + (3/30) = 37/120

Test Level 2: Probability - 2 - Question 20

If 8 coins are tossed, what is the chance that one and only one will turn up Head?

Detailed Solution for Test Level 2: Probability - 2 - Question 20

One head and seven tails would have eight positions where the head can come.
Thus, 8 x (1/2)8 = (1/32)

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