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This mock test of Test: Probability- 3 for Class 10 helps you for every Class 10 entrance exam.
This contains 25 Multiple Choice Questions for Class 10 Test: Probability- 3 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

If A be the event such that P(A) = 2/5, then P(not A) is equal to

Solution:

We know Probability law which is law of complement

Which is P(not A)=1-P(A)

Where P(A)=⅖

P(not A)=1-⅖=3/5

QUESTION: 2

What are the chances that no two boys are sitting together for a photograph if there are 5 girls and 2 boys?

Solution:

QUESTION: 3

In a single throw of a die, the probability of getting a multiple of 3 is

Solution:

Given : A die is thrown once .

A die has 6 faces marked as 1, 2, 3, 4, 5 and 6.

If we throw one die then there possible outcomes are as follows: 1, 2, 3, 4, 5 and 6

Number of possible outcomes are = 6

Let E = Event of getting a getting a multiple of 3

Multiples of 3 are = 3, 6

Number of outcome favourable to E = 2

Probability (E) = Number of favourable outcomes / Total number of outcomes

P(E) = 2/6 = 1/3

Hence, the probability of getting a multiple of 3, P(E) = 1/3

QUESTION: 4

Find the probability of getting a number greater than 2 when a die is thrown

Solution:

Probability of getting a number greater than 2 is

number greater than 2 = 4

total outcomes = 6

so, 4/6

=2/3

QUESTION: 5

The probability of getting a number between 1 and 6 is

Solution:

QUESTION: 6

Solution:

QUESTION: 7

The probability of getting one head is

Solution:

QUESTION: 8

What is the probability of getting two heads when a coin is tossed twice?

Solution:

QUESTION: 9

What is the probability of getting no head when two coins are tossed?

Solution:

Two coins are tossed simultaneously, so there are four possible outcomes ie, HH,HT,TT,TH

Total number of outcomes=4

Probability of getting no head=no. of favourable outcomes/ total no. of outcomes

=1/4

QUESTION: 10

In a simultaneous throw of two coins the probability of getting at least one head is

Solution:

Two coins are simultaneously tossed.

So sample space={HH,HT,TH,TT}

No. of favourable outcomes=getting at least one head={HH,HT,TH}

=3

Total number of outcomes=4

Probability of getting at least one head=No. of favourable outcomes/Total number of outcomes

=3/4

QUESTION: 11

The probability of getting a jack card is

Solution:

Total number of outcomes = 52

Favourable outcomes = four cards are of Jack.

Total number of favourable outcomes = 4

Probability = Number of favourable outcomes/ Total number of outcomes.

Required probability = P(jack) = 4/52= 1/13.

Hence, the probability of getting Jack is 1/13.

QUESTION: 12

The probability of getting a face card is

Solution:

Total Cards: 52

Total Face cards: 12 (4 Jacks, 4 Queens, 4 Kings)

P(Choosing a Face card) =12/52= 3/13

QUESTION: 13

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is :

The probability of getting a '10' of black suit is

Solution:

Let G be the favourable outcomes of getting 10 of black suit then

⇒ n(G) = 2

Therefore, P(G)

QUESTION: 14

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is :

The probability of getting a red and a king card is

Solution:

Let D be the favourable outcomes of getting red and a king, then n(D)=2 Therefore,

QUESTION: 15

A bag contains 4 red balls and 3 green balls. A ball is drawn at random. The probability of drawing a green ball is

Solution:

Total number of outcomes=7

No. of favourable outcomes=3 (no. Of green balls)

Probability of getting a green ball=No. Of favourable outcomes/Total no. of outcomes=3/7

QUESTION: 16

P(E) + P(E)¯ is equal to

Solution:

QUESTION: 17

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

Solution:

QUESTION: 18

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

Solution:

Probability of an event can only be a proper fraction and not an improper fraction.i.e.the numerator cannot be greater than the denominator.

P(E)=Number of outcomes favourable to E/Number of all possible outcomes of the experiment.

QUESTION: 19

Probability of an impossible event is equal to

Solution:

QUESTION: 20

If P(E_{1}) = 1/6, P (E_{2}) = 1/3 , P (E_{3}) = 1/6 , where E_{1}, E_{2}, E_{3} and E_{4} are elementary events of a random experiment, then P (E_{4}) is equal to

Solution:

QUESTION: 21

Cards each marked with one of the numbers 4, 5, 6,...., 20 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Then, the probability of getting an even prime number is

Solution:

We have one and only even prime number which is 2. So no. of favourable outcomes= 0 and total no. of outcomes=17

Probability of getting 2 =0 this means that 2 cannot occur .

QUESTION: 22

A bag contains 5 red and 4 black balls. A ball is drawn at random from the bag. Then, the probability of getting a black ball is

Solution:

QUESTION: 23

Find the probability of getting a head in a throw of a coin.

Solution:

QUESTION: 24

Two fair coins are tossed simultaneously. Find the probability of Getting only one head

Solution:

**he correct answer is a**

**When two coins are tossed, the sample space of possible outcomes would be the set**

**S = {HH, HT, TH, TT}**

**where H: the outcome is a “Head” and T: the outcome is a “Tail” and the probabilities associated with each of these outcomes are equal assuming the coins are fair.**

**When we say at most 1 “Head” outcome, it means that only 1 of the two coins show up a head or that the no “Head” outcome shows up.**

**If A is the event obtaining at most 1 “Head” outcome, then the set of outcomes that makes up event A is**

**A = {(HT), (TH), (TT)}**

**Hence, the probability of event A is**

**P(A) = 3/4 **

QUESTION: 25

Two fair coins are tossed simultaneously. Find the probability of Getting atleast one head

Solution:

When two coins are tossed simultaneously then the possible outcomes obtained are {HH, HT, TH, and TT}.

Here H denotes head and T denotes tail.

Therefore, a total of 4 outcomes obtained on tossing two coins simultaneously.

Consider the event of obtaining at the most one head. At most one head is obtained means either **no head** is obtained or **one head** is obtained.

The outcome(s) favourable to **no head** obtained is {TT}

The outcome(s) favourable to **one head** obtained is {HT, TH}

Therefore, the event of **obtaining at most one head** has 3 favourable outcomes. These are TT, HT and TH.

Therefore, the probability of obtaining at most one head = 3/4.

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