Test: Probability - 3 - Class 10 MCQ

# Test: Probability - 3 - Class 10 MCQ

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## 15 Questions MCQ Test The Complete SAT Course - Test: Probability - 3

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Test: Probability - 3 - Question 1

### If a coin is tossed thrice, find the probability of getting one or two heads.

Detailed Solution for Test: Probability - 3 - Question 1

Concept:

P(A) = n(A)/n(S)

Where n(A) = No. of favourable cases for event A and n(S) = cardinality of sample space.

Solution:

If a coin is tossed thrice, possible outcomes are:

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

Probability of getting one or two heads:

A = {HHT, HTH, THH, THT, TTH, HTT}

P(A) = 6/8

= 3/4

Test: Probability - 3 - Question 2

### What is the probability that a die is thrown? (1) A even number (2) A odd number (3) A prime number

Detailed Solution for Test: Probability - 3 - Question 2

We use the basic formula of probability to solve the problem.

The number of outcomes on throwing a die is (1, 2, 3, 4, 5, 6) = 6

(1) Total number of Even numbers are 2, 4 and 6 = 3

Probability of getting an even number = Number of Even Numbers/Total Number of Outcomes

⇒ 3/6 = 1/2

(2) Total number of Odd numbers are 1, 3 and 5 = 3

Probability of getting Odd number = Number of Odd Numbers/Total Number of Outcomes

⇒ 3/6 = 1/2

(3) Number of prime numbers on dice are 2, 3 and 5 = 3

Probability of getting a Prime number = Number of Prime Numbers/Total Number of Outcomes

⇒ 3/6 = 1/2

Test: Probability - 3 - Question 3

### A bag contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. The probability of getting the balls of different colors is:

Detailed Solution for Test: Probability - 3 - Question 3

Concept:

• The probability of drawing ‘k objects of type p’ from a collection of n = p + q + r + … objects is, given as:
• Probability of a Compound Event [(A and B) or (B and C)] is calculated as:

P[(A and B) or (B and C)] = [P(A) × P(B)] + [P(C) × P(D)]

('and' means '×' and 'or' means '+')

Calculation:

There are a total of 7 red + 4 blue = 11 balls.

Probability of drawing 1 red ball =

Probability of drawing 1 blue ball =

Probability of drawing (1 red) AND (1 blue) ball =

Similarly, Probability of drawing (1 blue) AND (1 red) ball =

Probability of getting the balls of different colors =

Test: Probability - 3 - Question 4

Suppose P(A) = 0.4, P(B) = P and P(A ∪ B) = 0.7. If A and B are independent events, then the value of P is:

Detailed Solution for Test: Probability - 3 - Question 4

Concept:

• For two events A and B, we have: P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
• If A and B are independent events, then P(A ∩ B) = P(A) × P(B).

Calculation:

Using the concept above, because A and B are independent events, we can write:

P(A ∪ B) = P(A) + P(B) - P(A) × P(B)

⇒ 0.7 = 0.4 + P - 0.4 × P

⇒ 0.6P =0.3

⇒ P = 0.5.

Test: Probability - 3 - Question 5

Three mangoes and three apples are in box. If two fruits are chosen at random, the probability that one is a mango and the other is an apple is

Detailed Solution for Test: Probability - 3 - Question 5

Concept:

If S is a sample space and E is a favourable event then the probability of E is given by:

P(E) = n(E)/n(S)

Calculation:

Total fruits = 3 + 3 = 6

Total possible ways = 6C2 = 15 = n(S)

Favourable ways = 3C1 × 3C1 = 9 = n(E)

∴ Required probability =

Test: Probability - 3 - Question 6

The number of possible outcomes, when a coin is tossed 6 times, is

Detailed Solution for Test: Probability - 3 - Question 6

Concept:

Sample space is nothing but a set of all possible outcomes of the experiment.

If we toss a coin n times then possible outcomes or number of elements in sample space = 2n elements

Calculation:

Number of outcomes when a coin is tossed = 2 (Head or Tail)

∴Total possible outcomes when a coin is tossed 6 times = 2 ×  2 × 2 × 2 × 2 × 2 = 64

Test: Probability - 3 - Question 7

Four dice are rolled. The number of possible outcomes in which at least one dice show 2 is

Detailed Solution for Test: Probability - 3 - Question 7

Concept:

Permutations with Repetition = nr

Where n is the number of things to choose from r different things when repetition is allowed, and order matters.

Favorable cases = Total cases - Unfavorable cases

Calculation:

According to the question

Four dies are rolled

So, Total Possible number of outcomes = 64

Now, Total outcomes when no 2 appears = 54

Now, From the concept used

Favorable cases = 64 - 54

⇒ 1296 - 625

⇒  671

∴ The number of possible outcomes in which at least one die shows 2 is 671.

Test: Probability - 3 - Question 8

One card is drawn out of a pack of 52 cards. What is the probability that the card drawn is a heart or a king?

Detailed Solution for Test: Probability - 3 - Question 8

Formula used:

1. Probability of occurrence of the event:

P(E) =  n(E)/n(S)

Where,

n(E) = Number of favorable outcome

n(S) = Number of possible outcome

2. P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Calculation:

Probability of choosing a heart

P(A) = 13/52

Probability of choosing a king

P(B) = 4/52

Probability of choosing king of heart

P(A ∩ B) = 1/52

We know that,

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

P(A ∪ B) = 13/52 + 4/52 - 1/52

P(A ∪ B) = (13+4-1)/51 = 16/52

P(A ∪ B) = 4/13

∴ The required probability will be 4/13.

Test: Probability - 3 - Question 9

If 2 dice are thrown what is the probability of getting the same digits on both dice?

Detailed Solution for Test: Probability - 3 - Question 9

When two dice are thrown together, then the total 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)       (6,6)}

∴ n(S) = 6 × 6 = 36

The favorable outcome of getting the same number on both dice is:

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).

∴ The number of favorable outcomes = 6

∴ Required probability = 6/36

= 1/6

Test: Probability - 3 - Question 10

If P(A ∪ B) = 3/4, P(A ∩ B) = 1/4, P(A̅ ) = 2/3, then find the value of P(B).

Detailed Solution for Test: Probability - 3 - Question 10

Concept:

P(A U B) = P(A) + P(B) - P(A  ∩ B)

Explanation:

P(A U B) = 3/4,

P(A ∩ B) = 1/4,

P( A̅ ) = 2/3

Since, P(A U B) = P(A) + P(B) - P(A  ∩ B)

Test: Probability - 3 - Question 11

A and B are two events such that P(B) = 0.4 and P(A ∪ B) = 0.6 If A and B are independent, then P(A) is

Detailed Solution for Test: Probability - 3 - Question 11

Concept:

Independent events:

Two events are independent if the incidence of one event does not affect the probability of the other event.

If A and B are two independent events, then P(A ∩ B) = P(A) × P(B)

Calculation:

Given: P(B) = 0.4 and P(A ∪ B) = 0.6

P(A ∪ B) = 0.6

⇒ P(A) + P(B) - P(A ∩ B) = 0.6

⇒ P(A) + P(B) - P(A) × P(B) = 0.6  (∵ A and B are independent events.)

⇒ P(B) + P(A) [1 - P(B)] = 0.6

⇒ 0.4 + P(A) [1 - 0.4] = 0.6

⇒ P(A) × 0.6 = 0.2

Test: Probability - 3 - Question 12

In a room there are eight couples. Out of them if 4 people are selected at random, the probability that they may be couples is

Detailed Solution for Test: Probability - 3 - Question 12

Concept:

1) Combination: Selecting r objects from given n objects.

• The number of selections of r objects from the given n objects is denoted by nCr

2) Probability of an event happening = Number of ways it can happen/Total number of outcomes

Note: Use combinations if a problem calls for the number of ways of selecting objects.

Calculation:

Given:

In a room, there are eight couples.

⇒ Eight couples = 16 peoples

We have to select four peoples out of 16 peoples.

⇒ Total possible cases = 16C4

Now, we have to select four people- they may be couples

So, we have to select two couples from eight couples.

⇒ Favourable cases = 8C2

Hence Required Probability =

Test: Probability - 3 - Question 13

An unbiased coin is tossed 3 times, if the third toss gets head what is the probability of getting at least one more head?

Detailed Solution for Test: Probability - 3 - Question 13

Concept:

• The number of ways for selecting r from a group of n (n > r) = nC
• The probability of particular case = Number of ways for the case can be executed/Total number of ways for selection

Calculation:

If it is known that third toss gets head, the possible cases:

(H, H, H), (H, T, H), (T, H, H), (T, T, H)

∴ Total cases possible = 4

Total favourable cases = 3 [(H, H, H), (H, T, H), (T, H, H)]

So, required probability P = Total favorable cases/Total possible cases

P = 3/4

Test: Probability - 3 - Question 14

If four dice are thrown together, then what is the probability that the sum of the numbers appearing on them is 25?

Detailed Solution for Test: Probability - 3 - Question 14

Concept:

Probability of an event happening = (Number of ways it can happen) / (Total number of outcomes)

If a die thrown, Number of sample space = 6, If two dice are thrown n(S) = 62 = 36

Calculation:

Here, four dice are thrown,

n(S) = 64

Now, sum of the numbers appearing on them 25 = { }

⇒ n = 0

(∵maximum sum = 6 + 6 + 6 + 6 = 24)

∴ Probability = 0/(64) = 0

Hence, option (a) is correct.

Test: Probability - 3 - Question 15

If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then

Detailed Solution for Test: Probability - 3 - Question 15

Concept:

• P(A|B) = P(A∩B)/P(B)
• P(B|A) = P(A∩B)/P(A)
• A ⊂ B = Proper Subset: every element of A is in B, but B has more elements.
• ϕ = Empty set = {}

Calculation:

Given: P(B/A) = 1

⇒ P(B|A) = P(A∩B)/P(A) = 1

⇒ P(A ∩ B) = P(A)

⇒ (A ∩ B) = A

So, every element of A is in B, but B has more elements.

∴ A ⊂ B

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