Probability Test

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Probability Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

In a single throw of a pair of dice, the probability of getting the sum as a perfect square is :

A
7/36

B
8/36

C
11/36

D
5/36

Solution

Total possible outcomes = 36

Number of favourable outcomes of getting the sum as a perfect square = { (1,3) , (2,2) ,(3,1) ,(3,6) ,(4,5) ,(5,4) ,(6,3)} = 7

Therefore, the probability of getting the sum as a perfect square = 7/36
Question 2
1 / -0

A number x is chosen at random from - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5. The probability that |x| ≤ 4 is :

A
9/11

B
6/11

C
7/11

D
3/11

Solution

Number of favourable outcomes = 11

Number of outcomes favourable to the event, |x| ≤ 4 = {-4, -3 ,-2, -1,0,1, 2, 3, 4} = 9

Therefore, the probability that |x| ≤ 4 = 9/11
Question 3
1 / -0

A die is thrown once. The probability of getting a number less than 3 is:

A
0

B
1

C
1/3

D
2/3

Solution

Total possible outcomes = 6

Total favourable outcomes of getting numbers less than 3 = {1,2} = 2

The probability of getting a number less than 3 = 2/6 = 1/3
Question 4
1 / -0

A die is thrown once. The probability of getting a number divisible by 2 is:

A
1/2

B
2/3

C
1

D
0

Solution

Total possible outcomes = 6

Number of favourable outcomes of gettinf a number divisible by 2 = {2,4,6} = 3

The probability of getting a number divisible by 2 = 3/6 = 1/2
Question 5
1 / -0

Which of the following cannot be the probability of an event ?

A
3%

B
0.1

C
17/16

D
1/3

Solution

The probability of an event lies between 0 and 1 (or 0-100% expressed in percentage). It cannot be greater than 1 and the value of 17/16 is greater than 1,so,it cannot be the probability of an event.
Question 6
1 / -0

A card is drawn at random from a pack of 52 cards, then the probability of getting a black ace is:

A
1/26

B
1/13

C
3/52

D
1/52

Solution

Total possible outcomes = 52

Number of favourable outcomes of getting black ace = {Ace of clubs and Ace of spades} = 2

Therefore, the probability of getting a black ace = 2/52 = 1/26
Question 7
1 / -0

A coin is tossed 3 times,then the probability of getting atleast two tails is:-

A
1/8

B
1/2

C
1/4

D
3/8

Solution

For three tosses of the coin all the possible outcomes are: {H-H-H, T-H-H, H-T-H, H-H-T, T-H-T, T-T-H, H-T-T, T-T-T } = 8

Favourable outcomes of getting atleast two tails ={ T-H-T, T-T-H, H-T-T, T-T-T} = 4

Therefore, the probability of getting atleast two tails = 4/8 = 1/2
Question 8
1 / -0

Three coins are tossed together, then the probability of getting atmost one head is:-

A
1

B
1/2

C
1/8

D
1/4

Solution

When three coins are tossed simultaneously, the total possible outcomes are {(H,H,H) , (H,H,T) , (H,T,H) , (H,T,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T)}.
The total number of outcomes = 8.

For atmost one head, number of heads should be either 0 or 1.
The favourable outcomes are ={(T,H,H), (T,H,T), (T,T,H), (T,T,T)}.
Number of favourable outcomes= 4.

Therefore, the probability of getting atmost one head = 4/8 = 1/2
Question 9
1 / -0

A bag contains 4 green, 5 black and 10 red balls.If one ball is drawn at random,the probability of not getting a green ball is

A
4/19

B
5/19

C
10/19

D
15/19

Solution

Total number of balls i.e. total possible outcomes = 4+5+10 =19

Total number of green balls = 4

Probability of getting a green ball P(E) = 4/19

Probability of not getting a green ball P(not E) = 1 - P(E) = 1 - 4/19 = 15/19
Question 10
1 / -0

Probability that a leap year, selected at random will contain 53 Saturdays is: -

A
53/365

B
7/365

C
1/7

D
2/7

Solution

A Leap year consists of 366 days.
So,we have, 52 weeks (52x7=364 days) and 2 days.
The extra 2 days may be:{(Sun,Mon),(Mon,Tue),(Tue,Wed),(Wed,Thu),(Thu,Fri),(Fri,Sat),(Sat,Sun)}
We have 7 options.
Among that, we have to choose only 2 options which has saturdays . (i.e, [Fri,Sat] or [Sat Sun])
So, the probability of having two Saturdays 2 out of 7 = 2/7

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Re-Attempt Weekly Quiz Competition

Probability Test - 9

Result

Probability Test - 9

Score

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Rank

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SHARING IS CARING

Question 1

7/36

8/36

11/36

5/36

Solution

Question 2

9/11

6/11

7/11

3/11

Solution

Question 3

0

1

1/3

2/3

Solution

Question 4

1/2

2/3

1

0

Solution

Question 5

3%

0.1

17/16

1/3

Solution

Question 6

1/26

1/13

3/52

1/52

Solution

Question 7

1/8

1/2

1/4

3/8

Solution

Question 8

1

1/2

1/8

1/4

Solution

Question 9

4/19

5/19

10/19

15/19

Solution

Question 10

53/365

7/365

1/7

2/7

Solution

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