# Free Probability 03 Practice Test - 9th Grade

### Question 1

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

#### 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.

### Question 2

The range of probability is

-1 to 1 (inclusive)

0 to 1 (inclusive)

1 to 2 (inclusive)

**-**1 to 0 (inclusive)

#### SOLUTION

Solution :B

.

### Question 3

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.

0.46

0.5

0.3

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

### Question 4

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?

1

0.67

0.25

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 4 = Frequency of occurrence of 1 or 4Total number of times die is thrownSo, probability of occurrence of 1 or 4 = 2080 = 14 = 0.25

### Question 5

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

13

0

1

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.

### Question 6

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

### Question 7

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

111

311

811

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.

### Question 8

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

0.5

1

0

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

### Question 9

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

#### 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

### Question 10

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

#### 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