# Free Probability 03 Practice Test - 9th Grade

When a coin is tossed, what is the probability of occurrence of "head" as the outcome?

A. 14
B. 35
C. 13
D. 12

#### SOLUTION

Solution : D

When you flip a coin, there are two outcomes either head or tail. So the probability of occurrence of the head is the same as the probability of occurrence of the tail that is 12

The range of probability is

A.

-1 to 1 (inclusive)

B.

0 to 1 (inclusive)

C.

1 to 2 (inclusive)

D.

-1 to 0 (inclusive)

#### SOLUTION

Solution : B

.

A coin was tossed 1500 times and out of these, 100 times, the coin stuck somewhere so the result was not counted. The frequency of occurrence of a head was 700. Find the probability of occurrence of a tail.

A.

0.46

B.

0.5

C.

0.3

D.

1

#### SOLUTION

Solution : B

A coin was tossed 1500 times but 100 times result was not counted.

So, the total number of times result counted = 1400

Number of times head comes = 700

Number of times tail comes = 1400 - 700 = 700

Probability of occurrence of tail =Number of times tail comesTotal number of times result counted

Probability of occurrence of tail =7001400

Probability of occurrence of tail = 0.5

Outcome2356Frequency10141620

A dice was thrown 80 times and the frequency of occurrence of 2, 3, 5 and 6 is given in the table. What is the probability of occurrence of 1 or 4?

A.

1

B.

0.67

C.

0.25

D.

14

#### SOLUTION

Solution : C and D

Total frequency of occurrence of 2, 3, 5 and 6 = 60.

Frequency of occurrence of 1 or 4 = 80 - 60 = 20.

Probability of occurrence of 1 or 4Frequency of occurrence of 1 or 4Total number of times die is thrown

So, probability of occurrence of 1 or 4 = 2080 = 14 = 0.25

A goldsmith can choose between 3 metals A, B or C to make an artifact. What is the probability that he chooses C?

A.

13

B.

0

C.

1

D.

23

#### SOLUTION

Solution : A

Since the goldsmith can choose amongst A, B and C, he has 3 possibilities of which the favorable outcome is choosing C. Therefore, probability that he chooses C is 13.

BagChocolates112219318422511

5 bags had some amount of chocolates in them, as shown in the table. The probability of occurrence of more than 18 chocolates in a bag if a bag is chosen at random will be ___

#### SOLUTION

Solution :

Out of the givne 5 bags, bag 2 and bag 4 have more than 18 chocolates.
Probability(E) =number of favorable eventstotal number of events
Probablity of a bag containing more than 18 chocolates =number of bags containing more than 18 chocolatestotal number of bags=25=0.4

If an alphabet is picked from the word MISSIMMIPPI what is the probability that an alphabet 'M' is picked?

A.

111

B.

311

C.

811

D.

1

#### SOLUTION

Solution : B

Total number of alphabets in the word MISSIMMIPPI = 11

Number of times M comes = 3

Probability of picking M = Number of times M comesTotal number of alphabets in the word MISSIMMIPPI

So, probability of its occurrence will be 311.

A coin was flipped 100 times. What is the maximum probability of occurrence of tail?

A.

0.5

B.

1

C.

0

D.

0.75

#### SOLUTION

Solution : B

A coin is tossed 100 times and consider the case when you only get tail 100 times and no head. So, in all the trials you will have a tail.

Probability of getting a tail = Number of times tails appearNumber of times a coin is tossed

Probability of getting a tail​ = 100100=1

What will be the probability that two friends have the same birthday date in a normal year ?

A. 173
B. 1365
C. 7365
D. 12365

#### SOLUTION

Solution : B

Number of days in a year = 365
Therefore, total possible outcomes = 365

If they have birthday on same day, then number of favourable outcomes =1

Therefore, required probability =1365

A die is rolled twice. Find the probability that 5 will come up both the times.

A. 136
B. 1836
C. 636
D. 436

#### SOLUTION

Solution : A

Number of all possible outcomes = (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) and (6,6).

The number of all possible outcomes = 36.

There is only one case (5,5) when 5 comes up both the times.

Probability that 5 will come up both the times=Number of times 5 comes up both the timesNumber of all possible outcomes

P (5 will come up both the times) =136