Probability - 2 - Free MCQ Practice Test with solutions, Class 10 The Complete

Given:

Two dices

Formula used:

Probability = No of desired outcomes/Total no of outcomes

Calculation:

We know that when 2 dice are thrown then total no of outcome will be 36.

According to question,

Outcomes with different numbers on dice = 30

Total no of outcomes = 36

Probability = 30/36

⇒ 5/6

∴ The correct answer is 5/6.

Given:

There are 3 fair coins.

Formula Used:

Probability of occurrence of an event = P(E) = Number of favorable outcomes/Number of possible outcomes = n(E)/n(S)

Calculation:

We know that:

When three fair coins are tossed simultaneously,

The numbers of possible outcomes are n(S)

n(S) = 2³ = 8 = {HHH, HHT, HTH, THH, TTT, TTH, THT, HTT} 

The number of favourable outcomes is

n(E) = {HTT, THT, TTH} = 3

Probability = 3/8

∴ The probability of getting one head = 3/8.

Given:

One dice shows a multiple of 3.

Other dice shows even number.

Concept:

Total number of outcomes in two dice is 36.

Formula used:

P = Favorable outcomes/Total outcomes 

Calculation:
There are only 6 such cases as required,

(3,2), (3,4) (3,6) (6,2) (6,4) (6,6)

∴ Required probability = 6/36 = 1/6

∴ The probability is 1/6.

Given:

Probability of passing the test by Sita = 60% = 60/100

Probability of passing the test by Gita = 40% = 40/100

Probability of passing the test by Mita = 20% = 20/100

Formula:

Probability of not happening even A = 1 - Probability of  happening even A

Probability of happening A and B = Probability of happening A × Probability of happening B

Calculation:

Probability of not passing the test by Mita = 1 - Probability of passing the test by Mita

= 1 - (20/100)

= 80/100

Now,

Probability that at Sita and Gita will pass the test and Mita will not = Probability of passing the test by Sita × Probability of passing the test by Gita × Probability of not passing the test by Mita

= (60/100) × (40/100) × (80/100)

= 192/1000

= (192/10)%

= 19.2%

Given:

A bag contains coins of one rupee, two rupee & five rupees.

The total money in the bag is Rs. 120.

The total number of one rupee and two rupee coins are 35 and in ratio of coins is 2 : 5.

Formula used:
Probability of an event = Favorable ways/Total ways

Calculation:

Number of 1 rupee coins = (2/7 × 35) = 10

Number of 2 rupee coins = (5/7 × 35) = 25

Let the number of 5 rupee coins be x

5x + 1(10) + 2(25) = 120 ⇒ x = 12

Total number of coins = 10 + 25 + 12 = 47

So the bag contain total 47 coins out of which number of five rupee coin is 12

Probability of getting a 5-rupee coin = 12/47

Given

Number of black balls = 5

Number of white balls = 6

Formula

Probability = Favorable events/Total possible events

Calculation

Favorable event = ⁶C₂

Total possible events = ¹¹C₂

∴ Probability = ⁶C₂/¹¹C₂ = (6 × 5)/(11 × 10) = 3/11

Given:

A team has six girls and six boys. Three students have to be selected for a project.

Calculation:

⇒ Total number of students = 12

⇒ Probability that three students are selected = ¹²C₃ = 12 × 11 × 10/6 = 220

⇒ Probability that two girls and one boy are selected = ⁶C₂ × ⁶C₁ = 6 × 5/2 × 6 = 90

∴ Required probability = 90/220 = 9/22

Given:

Three coins are tossed simultaneously.

Formula:

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

Calculation:

When three coins are tossed then the outcome will be any one of these combinations. (TTT, THT, TTH, THH. HTT, HHT, HTH, HHH).

So, the total number of outcomes is 8.

Now, for exactly two heads, the favorable outcome is (THH, HHT, HTH).

We can say that the total number of favorable outcomes is 3.

Again, from the formula

Probability = Number of favorable outcomes/Total number of outcomes

Probability = 3/8

∴ The probability of getting exactly two heads is 3/8.

Formula used:

n(A U B) = n(A) + n(B) - n(A ∩ B)

n(A) = Number of elements in A, n(B) = Number of elements in B

n(A U B) = Each element in both the set A and set B

n(A ∩ B) = The intersection contains the elements that the two sets have in common.

Calculation:

n(A) = 28

n(B) = 32

n(A U B) = 40

n(A ∩ B) = 32 + 28 - 40 = 20

Elements are present only in B = n(B) - n(A ∩ B) = 32 - 20 = 12

Given:

Number of bottle of juice = 3

Number of bottle of Lassi = 4

Number of bottle of milkshake = 6

Number of bottle picked up randomly = 2

Calculation:

Total number of bottles = (3 + 4 + 6) = 13

∴ Number of ways to choose 2 bottles = ¹³C₂ = 78

Number of lassi bottles = 4

∴ Number of ways to choose 2 lassi bottles = ⁴C₂ = 6

∴ Required probability = 6/78 = 1/13

Given:

No of possible outcomes when two dice are thrown simultaneously: 6 × 6 = 36 

(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), (6, 6)

Formula used:

Probability = No of favorable outcome ÷ No of total outcomes 

Calculation:

No of outcome with sum 12 (6, 6) = 1

∴ Required probability = 1/36 

Given:

Three students have to be chosen out of 12 students consisting of seven girls and five boys. Find the probability that at least one boy will be selected.

Formula:

Probability that at least one boy will be selected = 1 – (probability that no boys will be selected)

Calculation:

Probability that no boys will be selected = probability that only girls will be selected

⇒ Probability that only girls will be selected = ⁷C₃/¹²C₃ = 7 × 6 × 5/12 × 11 × 10 = 7/44

⇒ Probability that at least one boy will be selected = 1 – (7/44) = 37/44

∴ Required probability = 37/44

Formula Used: 

Probability = (Number of successful outcomes/Total number of outcomes)

P(E) = (nE)/(nS), where nE = Number of events and nS = Number of sample space 

Calculation:

When three coins are tossed, total possible outcomes = 8

S = {HHH, HHT, HTT, THH, TTH, THT, HTH}

Favourable cases = {HHT, HTT, THH, TTH, THT, HTH}

Here, H = Head, T = Tail

P(getting at least one head, one tail) = 6/8 = 3/4

∴ The probability is 3/4.

Given:

There are 32 new balls and 4 old balls in one bag.

Concept:

Number of ways to select ‘a’ out of ‘b’ = ^bC_a = b (b-1)/2  (when a = 2)

Probability = Number of favorable case / Total cases

Calculation:

32 new balls and 4 old balls in one bag:

Both the balls should be new.

∴, Required probability = ³²C₂ / ³⁶C₂

⇒ (32 × 31) / (36 × 35)

⇒ (8 × 31) / (9 × 35)

⇒ 248 / 315

Given:

The probability of passing an examination for A and B are 0.7 and 0.8 respectively. 

Formula Used:

Probability = Favorable outcome/sample space

Calculation:

⇒ Probability that A fail in examination = 1 - 0.7 = 0.3 

⇒ Probability that B fail in examination = 1 - 0.8 = 0.2 

⇒ required probability = 0.7 × 0.8 + 0.8 × 0.3 + 0.7 × 0.2 = 0.56 + 0.24 + 0.14 = 0.94 

∴ Required probability = 0.94 
Alternate Method:

Probability that at least one of them pass the examination = 1 - probability that both fails the examination

⇒ required probability = 1 - (0.3 × 0.2) = 0.94

∴ Required probability = 0.94