Test: Probability - 2 - Class 10 MCQ

# Test: Probability - 2 - Class 10 MCQ

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## 15 Questions MCQ Test The Complete SAT Course - Test: Probability - 2

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Test: Probability - 2 - Question 1

### Two dice are rolled simultaneously then the probability that they show different numbers is

Detailed Solution for Test: Probability - 2 - Question 1

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.

Test: Probability - 2 - Question 2

### Three fair coins are tossed simultaneously. Find the probability of getting one head.

Detailed Solution for Test: Probability - 2 - Question 2

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) = 2= 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.

Test: Probability - 2 - Question 3

### Ajay rolled two dice together. What is the probability that first dice showed a multiple of 3 and the second dice showed an even number?

Detailed Solution for Test: Probability - 2 - Question 3

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.

Test: Probability - 2 - Question 4

The probability of Sita, Gita and Mita passing a test is 60%, 40% and 20% respectively. What is the probability that at Sita and Gita will pass the test and Mita will not?

Detailed Solution for Test: Probability - 2 - Question 4

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%

Test: Probability - 2 - Question 5

A bag contains coins of one rupee, two rupee & five rupees. The total money in the bag is Rs. 120. If the total number of one rupee and two rupee coins are 35 and in ratio of coins is 2 : 5. Find the probability of getting a 5 rupee coin if a coin is randomly picked from the bag?

Detailed Solution for Test: Probability - 2 - Question 5

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

Test: Probability - 2 - Question 6

A bag contains 5 black and 6 white balls; two balls are drawn at random. What is the probability that the balls drawn are white?

Detailed Solution for Test: Probability - 2 - Question 6

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

Test: Probability - 2 - Question 7

A team has six girls and six boys. Three students have to be selected for a project. Find the probability that two girls and one boy are selected.

Detailed Solution for Test: Probability - 2 - Question 7

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

Test: Probability - 2 - Question 8

Three coins are tossed simultaneously. Find the probability of getting exactly two heads.

Detailed Solution for Test: Probability - 2 - Question 8

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.

Test: Probability - 2 - Question 9

Comprehension:

A has 28 elements, B has 32 elements and (A U B) has 40 elements.

How many elements are present only in B?

Detailed Solution for Test: Probability - 2 - Question 9

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

Test: Probability - 2 - Question 10

A box contains 3 bottles of juice, 4 bottles of lassi and 6 bottles of milkshake. 2 bottles are picked up randomly, what is the probability that both the bottles are of lassi?

Detailed Solution for Test: Probability - 2 - Question 10

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

Test: Probability - 2 - Question 11

Two dice are thrown simultaneously and the sum of the numbers appearing on them is noted. What is the probability that the sum is 12?

Detailed Solution for Test: Probability - 2 - Question 11

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

Test: Probability - 2 - Question 12

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.

Detailed Solution for Test: Probability - 2 - Question 12

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

Test: Probability - 2 - Question 13

Three fair coins are tossed simultaneously. What is the probability of getting at least one head and one tail?

Detailed Solution for Test: Probability - 2 - Question 13

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.

Test: Probability - 2 - Question 14

There are 32 new balls and 4 old balls in one bag. If any two balls are picked from the bag, what is the probability of both the balls being new?

Detailed Solution for Test: Probability - 2 - Question 14

Given:

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

Concept:

Number of ways to select ‘a’ out of ‘b’ = bCa = 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

Test: Probability - 2 - Question 15

The probability of passing an examination for A and B are 0.7 and 0.8 respectively. Find the probability that at least one of them pass the examination.

Detailed Solution for Test: Probability - 2 - Question 15

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

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