Probability Test

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Probability Test - 4

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Question 1
1 / -0

A sample space consists of numbers from 0 to 20. What is the probability that an element of the sample space is an even number, excluding numbers 0 and 20?

A
9/20

B
3/7

C
11/21

D
10/21

Solution

Sample space is 21.
Even numbers between 0 and 20 are {2, 4, 6, 8, 10, 12, 14, 16, 18}
Number of favourable cases = 9
Hence, the probability is 9/21 = 3/7.
Question 2
1 / -0

A single card is drawn from a well-shuffled deck of playing cards. What is the probability of the card being black with an even number?

A
5/26

B
5/52

C
2/13

D
1/4

Solution

There are total five even numbers (2, 4, 6, 8 and 10). So, there are total 10 black cards (club and spade) with an even number on them.
Total number of cards = 52

So, required probability = 10/52 = 5/26
Question 3
1 / -0

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

A
2/7

B
4/7

C
3/7

D
1/7

Solution

A leap year has 366 days, or 52 weeks + 2 days. This means that 52 Sundays will be there, but for the 53^rd, 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
Question 4
1 / -0

John, Mathew and Adam are playing with a pair of dice and have to obtain the sum of numbers 5, 6 and 7, respectively. Which of them has the highest possibility of winning?

A
John

B
Mathew

C
Adam

D
All have equal possibility

Solution

Probability of getting 5 is (2, 3), (3, 2), (1, 4) and (4, 1) i.e. 4/36 = 1/9
Probability of getting 6 is (2, 4), (4, 2), (3, 3), (5, 1) and (1, 5) i.e. 5/36

Similarly, probability of getting 7 is 6/36 = 1/6.
Thus, this is the highest amongst all of them.

Hence, Adam has the highest possibility of winning.