## RS Aggarwal Solutions Class 10 Chapter 15 Probability MCQ

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 15 Probability MCQ.

**Other Exercises**

- RS Aggarwal Solutions Class 10 Chapter 15 Probability Ex 15a
- RS Aggarwal Solutions Class 10 Chapter 15 Probability Ex 15b
- RS Aggarwal Solutions Class 10 Chapter 15 Probability MCQ

**Question 1.**

**Solution:
**Correct option: (c)

We know that, the probability of an event E will always lie between O and 1, where O is the probability of an impossible event and 1 is the probability of a sure event.

**Question 2.**

**Solution:
**Correct option: (b)

Let E be the event.

So, the probability of the event happening will be P(E).

Thus, the probability of the event not happening will be P(E’).

Given that, P(E)= p

We know that, P(E’)+ P(E)= 1

p + P(E’) = 1

=P(E’)= 1 – p

**Question 3.**

**Solution:
**Correct option: (b)

The probability of an impossible event is always O.

**Question 4.**

**Solution:
**Correct option: (c)

The probability of a sure event is always 1.

**Question 5.**

**Solution:
**Correct option: (a)

We know that, the probability of an event E will always lie between O and 1.

Since 1.5>1, it cannot be the probability of an event.

**Question 6.**

**Solution:
**

**Question 7.**

**Solution:
**

**Question 8.**

**Solution:
**

**Question 9.**

**Solution:
**

**Question 10.**

**Solution:
**Correct option: (d)

The total number of cards = 9

The primes numbers would be 2, 3, 5, 7, 11.

So, there are 5 numbers.

P(getting a prime number) = 5/9.

**Question 11.**

**Solution:
**Correct option: (b)

The total number of tickets = 40

The multiples of 7 between 1 and 40 are 7, 14, 21, 28 and 35.

So, there are 5 numbers,

P(getting a multiple of 7) = 5/40 = 1/8.

**Question 12.
**

**Solution:**

Correct option: (d)

We know that, the probability of an event E will always lie between O and 1.

Since 7/6 >1, it cannot be the probability of an event.

**Question 13.**

**Solution:
**

**Question 14.**

**Solution:
**Correct option: (d)

An event that cannot occur is called an impossible event.

The probability of an impossible event is 0.

**Question 15.**

**Solution:
**Correct option: (b)

The total number of tickets = 20

The multiples of 5 between 1 and 20 are 5, 10, 15 and 20.

So, there are 4 numbers.

P(getting a multiple of 5) = 4/20 = 1/5.

**Question 16.**

**Solution:
**

**Question 17.**

**Solution:
**Correct option: (d)

The total number of bckets = 10

The numbers less than 10 are 6, 7, 8 and 9.

So, there are 4 numbers.

P(getting a number less than 10) = 2/5.

**Question 18.**

**Solution:
**Correct option: (a)

The numbers on a die are 1, 2, 3, 4, 5 and 6.

So, there are 6 numbers in total.

The even numbers on the die are 2, 4 and 6.

So, there are 3 numbers.

P(getting an even number) = 3/6 = 1/2.

**Question 19.**

**Solution:
**Correct option: (d)

The numbers on a fair die are 1, 2, 3, 4, 5 and 6.

So, there are 6 numbers in total.

The numbers greater than 2 are 3, 4, 5 and 6.

So, there are 4 numbers.

P(getting a number greater than 2) = 4/6 = 2/3.

**Question 20.**

**Solution:
**Correct option: (b)

The numbers on a die are 1, 2, 3, 4, 5 and 6.

So, there are 6 numbers in total.

The odd number on a die greater than 3 is 5.

So, there is only 1 number.

P(getting an odd number greater than 3) = 1/6.

**Question 21.**

**Solution:
**Correct option: (c)

The numbers on a die are 1, 2, 3, 4, 5 and 6.

So, there are 6 numbers in total.

The prime numbers on the die are 2, 3 and 5.

So, there are 3 numbers.

P(getting a prime number on the die) = 3/6 = 1/2.

**Question 22.**

**Solution:
**

**Question 23.**

**Solution:
**Correct option: (d)

When two coins are tossed the outcomes are:

{HH, HT, TH, TT}

So, there are 4 numbers in total.

P(getting one head) = 1/4.

**Question 24.**

**Solution:
**

**Question 25.**

**Solution:
**Correct option: (d)

When two coins are tossed the simultaneously the outcomes are:

{HH, HT, TH, TT}

So, there are 4 outcomes.

Getting atmost one head means the possible outcomes are:

{HT, TH, TT}

So, there are 3 possible outcomes.

P(getting atmost one head) = 3/4.

**Question 26.**

**Solution:
**Correct option: (c)

When three coins are tossed the simultaneously the

outcomes are:

{HHH, HHT, HTH, THH, THT, HTT, TTH and TTT}

So, there are 8 possible outcomes.

P(getting exactly two heads) = 3/8.

**Question 27.**

**Solution:
**Correct option: (b)

The number of prizes = 8

The number of blanks = 16

So, the total number of tickets=8+ 16 = 24

P(getting a prize) = 8/24 = 1/3.

**Question 28.**

**Solution:
**Correct option: (c)

The number of prizes= 6

The number of blanks = 24

So, the total number of tickets = 6+24=30

P(not getting a prize)= 24/30 = 4/5.

**Question 29.**

**Solution:
**Correct option: (c)

The bag contains 3 blue, 2 white and 4 red marbles.

So, the total number of marbles = 3+2+4= 9

Since the marbles cannot be white, it can blue or red.

The number of blue or red marbles= 3+4=7

P(getting a blue or red marble)= 7/9.

**Question 30.**

**Solution:
**Correct option: (b)

The bag contains 4 red and 6 black balls.

So, the total number of balls= 4+6= 10

The number of black balls = 6

P(getting a black ball) = 6/10 = 3/5.

**Question 31.**

**Solution:
**Correct option: (c)

The bag contains 8 red, 2 black and 5 white balls,

So, the total number of balls= 8+2+5=15

Since the ball should not be black, it can be red or white.

The number of red and white balls = 13

P(getting a red and white ball) = 13/15.

**Question 32.**

**Solution:
**Correct option: (c)

The bag contains 3 white, 4 red and 5 black balls.

So, the total number of balls = 3+ 4+ 5= 12

For the ball that is drawn to be neither black nor white, it should be red.

The number of red balls = 4

P(getting a red ball) = 4/12 = 1/3.

**Question 33.**

**Solution:
**Correct option: (b)

The total number of cards = 52

The number of black kings = 2

P(getting a black king) = 2/52 = 1/26.

**Question 34.**

**Solution:
**Correct option: (a)

The total number of cards = 52

The number of queens = 4

P(getting a queen) = 4/52 = 1/13.

**Question 35.**

**Solution:
**Correct option: (c)

The total number of cards = 52

The number of face cards = 12

P(getting a face card) = 12/ 52 = 3/13.

**Question 36.**

**Solution:
**Correct option: (b)

The total number of cards = 52

The number of black face cards = 6

P(getting a black face card) = 6/52 = 3/26.

**Question 37.**

**Solution:
**Correct option: (c)

The total number of cards = 52

The number of 6 in the deck of cards = 4

P(getting a 6) = 4/52 = 1/13.

Hope given RS Aggarwal Solutions Class 10 Chapter 15 Probability MCQ are helpful to complete your math homework.

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