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Maths Probability Test.18

**Quiz Instructions:**

- There will be 20 multiple choice question in this online test.
- Answer of the questions will change randomly each time you start this test.
- Practice this test at least 3 times if you want to secure High Marks.
- At the End of the Test you can see your Test score and Rating.

1 / 20

**In a drawer there are 4 white socks, 3 blue socks and 5 grey socks. Two socks are picked randomly. What is the possibility that both the socks are of same color?**

2 / 20

**What is probability of drawing two clubs from a well shuffled pack of 52 cards?**

3 / 20

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

4 / 20

**A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is___________?**

5 / 20

**From a pack of cards two cards are drawn one after the other, with replacement. The probability that the first is a red card and the second is a king is_________?**

Explanation:

Let E1 be the event of drawing a red card.

Let E2 be the event of drawing a king .

P(E1 ∩ E2) = P(E1) . P(E2)

(As E1 and E2 are independent)

= 1/2 * 1/13 = 1/26

Explanation:

Let E1 be the event of drawing a red card.

Let E2 be the event of drawing a king .

P(E1 ∩ E2) = P(E1) . P(E2)

(As E1 and E2 are independent)

= 1/2 * 1/13 = 1/26

6 / 20

**Out of 10 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is at least one graduate among them?**

Explanation:

P(at least one graduate) = 1 – P(no graduates) =

1 – ⁶C₃/¹⁰C₃ = 1 – (6 * 5 * 4)/(10 * 9 * 8) = 5/6

Explanation:

P(at least one graduate) = 1 – P(no graduates) =

1 – ⁶C₃/¹⁰C₃ = 1 – (6 * 5 * 4)/(10 * 9 * 8) = 5/6

7 / 20

**The probability that A speaks truth is 3/5 and that of B speaking truth is 4/7. What is the probability that they agree in stating the same fact?**

Explanation:

If both agree stating the same fact, either both of them speak truth of both speak false.

Probability = 3/5 * 4/7 + 2/5 * 3/7

= 12/35 + 6/35 = 18/35

Explanation:

If both agree stating the same fact, either both of them speak truth of both speak false.

Probability = 3/5 * 4/7 + 2/5 * 3/7

= 12/35 + 6/35 = 18/35

8 / 20

**In a party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples?**

Explanation:

Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ * ⁶C₁)

As remaining person can be any one among three couples left.

Required probability = (⁵C₂ * ⁶C₁)/¹⁰C₅

= (10 * 6)/252 = 5/21

Explanation:

Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ * ⁶C₁)

As remaining person can be any one among three couples left.

Required probability = (⁵C₂ * ⁶C₁)/¹⁰C₅

= (10 * 6)/252 = 5/21

9 / 20

**The probability of a lottery ticket being a prized ticket is 0.2. When 4 tickets are purchased, the probability of winning a prize on at least one ticket is__________?**

Explanation:

P(winning prize atleast on one ticket)

= 1 – P(“Losing on all tickets”)

= 1 – (0.8)4 = (1 + (0.8)2)(1 – (0.8)2)

= (1.64)(0.36) = 0.5904

Explanation:

P(winning prize atleast on one ticket)

= 1 – P(“Losing on all tickets”)

= 1 – (0.8)4 = (1 + (0.8)2)(1 – (0.8)2)

= (1.64)(0.36) = 0.5904

10 / 20

**10 books are placed at random in a shelf. The probability that a pair of books will always be together is_________?**

Explanation:

10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2!

Required probability = 1/5

Explanation:

10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2!

Required probability = 1/5

11 / 20

**What is the probability that a leap year has 53 Sundays and 52 Mondays?**

Explanation:

A leap year has 52 weeks and two days

Total number of cases = 7

Number of favourable cases = 1

i.e., {Saturday, Sunday}

Required Probability = 1/7

Explanation:

A leap year has 52 weeks and two days

Total number of cases = 7

Number of favourable cases = 1

i.e., {Saturday, Sunday}

Required Probability = 1/7

12 / 20

**Out of 15 consecutive numbers, 2 are chosen at random. The probability that they are both odds or both primes is_________?**

Explanation:

There is no definite formula for finding prime numbers among 15 consecutive numbers. Hence the probability cannot be determined.

Explanation:

There is no definite formula for finding prime numbers among 15 consecutive numbers. Hence the probability cannot be determined.

13 / 20

**A bag contains five white and four red balls. Two balls are picked at random from the bag. What is the probability that they both are different color?**

Explanation:

Two balls can be picked from nine balls in ⁹C₂ ways.

We select one white ball and one red ball from five white balls and four red balls. This can be done ⁵C₁ . ⁴C₁ ways.

The required probability = (5 * 4)/⁹C₂ = 20/36 = 5/9

Explanation:

Two balls can be picked from nine balls in ⁹C₂ ways.

We select one white ball and one red ball from five white balls and four red balls. This can be done ⁵C₁ . ⁴C₁ ways.

The required probability = (5 * 4)/⁹C₂ = 20/36 = 5/9

14 / 20

**A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are picked at random, what is the probability that they are either blue or yellow?**

Explanation:

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. Probability that both marbles are blue = ³C₂/¹⁵C₂ = (3 * 2)/(15 * 14) = 1/35

Probability that both are yellow = ²C₂/¹⁵C₂ = (2 * 1)/(15 * 14) = 1/105

Probability that one blue and other is yellow = (³C₁ * ²C₁)/¹⁵C₂ = (2 * 3 * 2)/(15 * 14) = 2/35

Required probability = 1/35 + 1/105 + 2/35

= 3/35 + 1/105 = 1/35(3 + 1/3)

= 10/(3 * 35) = 2/21

Explanation:

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. Probability that both marbles are blue = ³C₂/¹⁵C₂ = (3 * 2)/(15 * 14) = 1/35

Probability that both are yellow = ²C₂/¹⁵C₂ = (2 * 1)/(15 * 14) = 1/105

Probability that one blue and other is yellow = (³C₁ * ²C₁)/¹⁵C₂ = (2 * 3 * 2)/(15 * 14) = 2/35

Required probability = 1/35 + 1/105 + 2/35

= 3/35 + 1/105 = 1/35(3 + 1/3)

= 10/(3 * 35) = 2/21

15 / 20

**A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If three marbles are picked at random, what is the probability that they are all blue?**

Explanation:

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.

Probability that all the three marbles picked at random are blue = ³C₃/¹⁵C₃ = (1 * 3 * 2 * 1)/(15 * 14 * 13) = 1/455

Explanation:

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.

Probability that all the three marbles picked at random are blue = ³C₃/¹⁵C₃ = (1 * 3 * 2 * 1)/(15 * 14 * 13) = 1/455

16 / 20

**A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If four marbles are picked at random, what is the probability that none is blue?**

Explanation:

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. When four marbles are picked at random, then the probability that none is blue is = ¹²C₄/¹⁵C₄ = (12 * 11 * 10 * 9)/(15 * 14 * 13 * 12) = 33/91

Explanation:

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. When four marbles are picked at random, then the probability that none is blue is = ¹²C₄/¹⁵C₄ = (12 * 11 * 10 * 9)/(15 * 14 * 13 * 12) = 33/91

17 / 20

**A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that at least one bulb is good.**

Explanation:

Required probability = 1 – 1/126 = 125/126

Explanation:

Required probability = 1 – 1/126 = 125/126

18 / 20

**A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that all the four bulbs are defective.**

Explanation:

Out of nine, five are good and four are defective. Required probability = ⁴C₄/⁹C₄ = 1/126

Explanation:

Out of nine, five are good and four are defective. Required probability = ⁴C₄/⁹C₄ = 1/126

19 / 20

**Three 6 faced dice are thrown together. The probability that exactly two dice show the same number on them is -.**

Explanation:

Using question number 11 and 12, we get the probability as

1 – (1/36 + 5/9) = 5/12

Explanation:

Using question number 11 and 12, we get the probability as

1 – (1/36 + 5/9) = 5/12

20 / 20

**Three 6 faced dice are thrown together. The probability that no two dice show the same number on them is__________?**

Explanation:

No two dice show same number would mean all the three faces should show different numbers. The first can fall in any one of the six ways. The second die can show a different number in five ways. The third should show a number that is different from the first and second. This can happen in four ways.

Thus 6 * 5 * 4 = 120 favourable cases.

The total cases are 6 * 6 * 6 = 216.

The probability = 120/216 = 5/9.

Explanation:

No two dice show same number would mean all the three faces should show different numbers. The first can fall in any one of the six ways. The second die can show a different number in five ways. The third should show a number that is different from the first and second. This can happen in four ways.

Thus 6 * 5 * 4 = 120 favourable cases.

The total cases are 6 * 6 * 6 = 216.

The probability = 120/216 = 5/9.