Probability Class 10 Important Questions PYQ

Given

A bag contains:

• Black balls = 7
• Red balls = 5
• White balls = 3

Total balls = 7 + 5 + 3 = 15

Probability formula:

$$\text{Probability} = \frac{\text{Favourable outcomes}}{\text{Total outcomes}}$$

1. Probability of drawing a red ball

Favourable outcomes = 5

$$P(\text{Red})=\frac{5}{15}=\frac{1}{3}$$

Answer: $\boxed{\tfrac{1}{3}}$

2. Probability of drawing a black or white ball

Black or White balls = 7 + 3 = 10

$$P(\text{Black or White})=\frac{10}{15}=\frac{2}{3}$$

Answer: $\boxed{\tfrac{2}{3}}$

3. Probability of drawing a ball that is not black

Not black = Red + White = 5 + 3 = 8

$$P(\text{Not Black})=\frac{8}{15}$$

Answer: $\boxed{\tfrac{8}{15}}$

Given

• Red balls = 7
• White balls = 5
• Black balls = 3

Total balls = 7 + 5 + 3 = 15

Formula: Probability = Favourable outcomes ÷ Total outcomes

1. Probability of drawing a red or white ball

Red or White balls = 7 + 5 = 12

$$P(\text{Red or White})=\frac{12}{15}=\frac{4}{5}$$

Answer: $\boxed{\frac{4}{5}}$

2. Probability of drawing a ball that is not black

Not black = Red + White = 7 + 5 = 12

$$P(\text{Not Black})=\frac{12}{15}=\frac{4}{5}$$

Answer: $\boxed{\frac{4}{5}}$

3. Probability of drawing a ball that is neither white nor black

“Neither white nor black” means red only.

Red balls = 7

$$P(\text{Neither White nor Black})=\frac{7}{15}$$

Answer: $\boxed{\frac{7}{15}}$

Given

• Red balls = 5
• Blue balls = 4
• Green balls = 3

Total balls = 5 + 4 + 3 = 12

Probability = Favourable outcomes ÷ Total outcomes

1. Probability of selecting a red ball

Favourable outcomes = 5

$$P(\text{Red})=\frac{5}{12}$$

Answer: $\boxed{\frac{5}{12}}$

2. Probability of selecting a ball that is not green

Not green = Red + Blue = 5 + 4 = 9

$$P(\text{Not Green})=\frac{9}{12}=\frac{3}{4}$$

Answer: $\boxed{\frac{3}{4}}$

Given

Two dice are thrown together.

• Total number of outcomes = 6 × 6 = 36

Sample space = 36 outcomes

1. Probability that 5 will not come up on either die

A 5 should not appear on both dice.

On one die, the possible numbers are: 1, 2, 3, 4, 6 (5 outcomes)

So, favourable outcomes = 5 × 5 = 25

$$P=\frac{25}{36}$$

Answer: $\boxed{\frac{25}{36}}$

2. Probability that 5 will come up on at least one die

“At least one 5” means 5 appears on the first die, or second die, or both.

Favourable outcomes:

• First die = 5 → 6 outcomes
• Second die = 5 → 6 outcomes
• (5,5) counted twice, so subtract 1

Total favourable outcomes = 6 + 6 − 1 = 11

$$P=\frac{11}{36}$$

Answer: $\boxed{\frac{11}{36}}$

3. Probability that 5 will come up on both dice

Only one outcome satisfies this: (5,5)

Favourable outcomes = 1

$$P=\frac{1}{36}$$

Answer: $\boxed{\frac{1}{36}}$

Given

A box contains 25 cards numbered from 1 to 25.

• Total number of cards = 25

Formula: Probability = Favourable outcomes / Total outcomes

1. Probability of drawing an even number

Even numbers from 1 to 25 are:

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24

Number of even cards = 12

$$P(\text{Even})=\frac{12}{25}$$

Answer: $\boxed{\frac{12}{25}}$

2. Probability of drawing a prime number

Prime numbers from 1 to 25 are:

2, 3, 5, 7, 11, 13, 17, 19, 23

Number of prime cards = 9

$$P(\text{Prime})=\frac{9}{25}$$

Answer: $\boxed{\frac{9}{25}}$

3. Probability of drawing a multiple of 6

Multiples of 6 from 1 to 25 are:

6, 12, 18, 24

Number of multiples of 6 = 4

$$P(\text{Multiple of 6})=\frac{4}{25}$$

Answer: $\boxed{\frac{4}{25}}$

Given

A box contains 19 balls numbered from 1 to 19.

• Total number of balls = 19

Formula: $\text{Probability} = \frac{\text{Favourable outcomes}}{\text{Total outcomes}}$

1. Probability of drawing a prime number

Prime numbers from 1 to 19 are:

2, 3, 5, 7, 11, 13, 17, 19

Number of prime numbers = 8

$$P(\text{Prime})=\frac{8}{19}$$

Answer: $\boxed{\frac{8}{19}}$

2. Probability of drawing a number divisible by 3 or 5

Numbers divisible by 3 or 5 are:

• Divisible by 3: 3, 6, 9, 12, 15, 18
• Divisible by 5: 5, 10, 15

Common number = 15 (count only once)

Favourable numbers = 3, 5, 6, 9, 10, 12, 15, 18

Total favourable outcomes = 8

$$P=\frac{8}{19}$$

Answer: $\boxed{\frac{8}{19}}$198​​

3. Probability of drawing a number neither divisible by 5 nor by 10

Numbers divisible by 5 or 10 are:

5, 10, 15

So, numbers neither divisible by 5 nor by 10 = 19 − 3 = 16

$$P=\frac{16}{19}$$

Answer: $\boxed{\frac{16}{19}}$

4. Probability of drawing an even number

Even numbers from 1 to 19 are:

2, 4, 6, 8, 10, 12, 14, 16, 18

Number of even numbers = 9

$$P(\text{Even})=\frac{9}{19}$$

Answer: $\boxed{\frac{9}{19}}$

Given

A box contains 20 balls numbered from 1 to 20.

• Total number of balls = 20

Formula:

$$\text{Probability} = \frac{\text{Favourable Outcomes}}{\text{Total Outcomes}}$$

1. Probability of drawing an odd number

Odd numbers from 1 to 20 are:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Number of odd numbers = 10

$$P(\text{Odd})=\frac{10}{20}=\frac{1}{2}$$

Answer: $\boxed{\frac{1}{2}}$

2. Probability of drawing a number divisible by 2 or 3

Numbers divisible by 2:

2, 4, 6, 8, 10, 12, 14, 16, 18, 20 (10 numbers)

Numbers divisible by 3:

3, 6, 9, 12, 15, 18 (6 numbers)

Common numbers (divisible by both 2 and 3):

6, 12, 18 (3 numbers)

So, favourable outcomes:

10 + 6 − 3 = 13

$$P=\frac{13}{20}$$

Answer: $\boxed{\frac{13}{20}}$

3. Probability of drawing a prime number

Prime numbers from 1 to 20 are:

2, 3, 5, 7, 11, 13, 17, 19

Number of prime numbers = 8

$$P(\text{Prime})=\frac{8}{20}=\frac{2}{5}$$

Answer: $\boxed{\frac{2}{5}}$

4. Probability of drawing a number not divisible by 10

Numbers divisible by 10:

10, 20

Number of such balls = 2

Numbers not divisible by 10 = 20 − 2 = 18

$$P=\frac{18}{20}=\frac{9}{10}$$

Answer: $\boxed{\frac{9}{10}}$

Step 1: Total number of days in a leap year

A leap year has 366 days.

$$366=52\text{ weeks}+2\text{ days}$$366=52 weeks+2 days

So, every weekday occurs 52 times, and the remaining 2 days occur 53 times.

Step 2: Possible extra days

The two extra days can be:

1. Monday & Tuesday
2. Tuesday & Wednesday
3. Wednesday & Thursday
4. Thursday & Friday
5. Friday & Saturday
6. Saturday & Sunday
7. Sunday & Monday

There are 7 equally likely possibilities.

Step 3: Favourable outcomes

A leap year has 53 Fridays if Friday is one of the two extra days.

This happens in 2 cases:

• Thursday & Friday
• Friday & Saturday

Favourable outcomes = 2

Step 4: Probability

$$P(53\text{ Fridays})=\frac{2}{7}$$

Final Answer

$$\boxed{\frac{2}{7}}$$

Probability of getting 53 Fridays in a leap year = $\frac{2}{7}$.

Given

• Probability of winning = 0.6

We know:

$$P(\text{Winning})+P(\text{Losing})=1$$

Calculation

$$P(\text{Losing})=1-0.6=0.4$$

Final Answer

Probability of losing the game = $\boxed{0.4}$​ or $\boxed{\tfrac{2}{5}}$

Given: A well-shuffled deck has 52 cards.

Formula:

$$\text{Probability} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$$

1. Probability of drawing an ace

• Number of aces = 4

$$P(\text{Ace})=\frac{4}{52}=\frac{1}{13}$$

Answer: $\boxed{\frac{1}{13}}$

2. Probability of drawing the 4 of spades

• Only one 4 of spades

$$P(\text{4 of Spades})=\frac{1}{52}$$

Answer: $\boxed{\frac{1}{52}}$

3. Probability of drawing a 9 of a black suit

Black suits are Spades and Clubs.

• Black 9s = 9♠ and 9♣ = 2 cards

$$P(\text{9 of a black suit})=\frac{2}{52}=\frac{1}{26}$$

Answer: $\boxed{\frac{1}{26}}$

4. Probability of drawing a red king

Red suits are Hearts and Diamonds.

• Red kings = 2

$$P(\text{Red King})=\frac{2}{52}=\frac{1}{26}$$

Answer: $\boxed{\frac{1}{26}}$

Given

A well-shuffled deck has 52 playing cards.

• Total number of cards = 52
• Probability = Favourable outcomes / Total outcomes

1. Probability of drawing an Ace

There are 4 Aces in a deck.

$$P(\text{Ace})=\frac{4}{52}=\frac{1}{13}$$

Answer: $\boxed{\frac{1}{13}}$

2. Probability of drawing the 4 of Spades

There is only one 4 of Spades.

$$P(4\text{ of Spades})=\frac{1}{52}$$

Answer: $\boxed{\frac{1}{52}}$

3. Probability of drawing a 9 of a black suit

Black suits are Spades and Clubs.

The black 9s are:

• 9 of Spades
• 9 of Clubs

Favourable cards = 2

$$P(\text{Black 9})=\frac{2}{52}=\frac{1}{26}$$

Answer: $\boxed{\frac{1}{26}}$

4. Probability of drawing a Red King

Red suits are Hearts and Diamonds.

Red Kings are:

• King of Hearts
• King of Diamonds

Favourable cards = 2

$$P(\text{Red King})=\frac{2}{52}=\frac{1}{26}$$

Answer: $\boxed{\frac{1}{26}}$

Given

Given

• Total number of cards = 52
• Red cards = 26
• Queens = 4
• Red queens = 2 (Queen of Hearts and Queen of Diamonds)

Step 1: Find the cards that are red or queens

Since the 2 red queens are counted twice, subtract them once.

$$26+4-2=28$$

So, 28 cards are either red or queens.

Step 2: Find the cards that are neither red nor queens

$$52-28=24$$

Favourable outcomes = 24

Step 3: Find the probability

$$P=\frac{24}{52}=\frac{6}{13}$$

Final Answer

$$\boxed{\frac{6}{13}}$$136​​

Solution

Total number of cards = 52

Queens in a pack = 4

Jacks in a pack = 4

Therefore, cards that are queens or jacks = 4 + 4 = 8

Cards that are neither queen nor jack:

$$52-8=44$$

Probability

$$P(\text{Neither Queen nor Jack})=\frac{44}{52}=\frac{11}{13}$$

A well-shuffled deck contains 52 playing cards.

• Spades = 13 cards
• Hearts = 13 cards
• Diamonds = 13 cards
• Clubs = 13 cards
• Kings = 4 cards
• Queens = 4 cards
• Aces = 4 cards

Formula: Probability = Favourable Outcomes / Total Outcomes

Probability of drawing a spade or an ace

• Number of spades = 13
• Number of aces = 4
• Common card (Ace of Spades) = 1

$$\text{Favourable outcomes}=13+4-1=16$$

$$P(\text{Spade or Ace})=\frac{16}{52}=\frac{4}{13}$$

Answer: $\boxed{\frac{4}{13}}$

2. Probability of drawing a red king

Red suits are Hearts and Diamonds.

Red kings are:

• King of Hearts
• King of Diamonds

Favourable outcomes = 2

$$P(\text{Red King})=\frac{2}{52}=\frac{1}{26}$$

Answer: $\boxed{\frac{1}{26}}$

3. Probability of drawing either a king or a queen

• Kings = 4
• Queens = 4

A card cannot be both a king and a queen.

Favourable outcomes = 4 + 4 = 8

$$P(\text{King or Queen})=\frac{8}{52}=\frac{2}{13}$$

Answer: $\boxed{\frac{2}{13}}$

4. Probability of drawing neither a king nor a queen

Cards that are kings or queens = 8

Cards that are neither = 52 − 8 = 44

$$P(\text{Neither King nor Queen})=\frac{44}{52}=\frac{11}{13}$$

Answer: $\boxed{\frac{11}{13}}$

Given:

• Total cards in a pack = 52
• Lost cards = King, Queen, Jack and 10 of Spades = 4 cards
• Remaining cards = 52 − 4 = 48

Formula:

$$\text{Probability} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$$

1. Probability of getting a red card

All red cards are still present.

• Red cards = 26

$$P(\text{Red card})=\frac{26}{48}=\frac{13}{24}$$

Answer: $\boxed{\frac{13}{24}}$

2. Probability of getting a king

There are 4 kings originally, but the King of Spades is lost.

• Kings left = 3

$$P(\text{King})=\frac{3}{48}=\frac{1}{16}$$

Answer: $\boxed{\frac{1}{16}}$

3. Probability of getting a black card

Originally black cards = 26

Four spade cards are lost, so:

• Black cards left = 26 − 4 = 22

$$P(\text{Black card})=\frac{22}{48}=\frac{11}{24}$$

Answer: $\boxed{\frac{11}{24}}$

A standard deck has 52 cards.

Removed cards: Red Kings, Red Queens, and Red Jacks

• Red Kings = 2
• Red Queens = 2
• Red Jacks = 2

Total removed = 6 cards

So, remaining cards = 52 − 6 = 46

Probability = $\frac{\text{Favourable outcomes}}{46}$

1. Probability of getting a king

Originally there are 4 kings. Two red kings are removed, so 2 black kings remain.

$$P(\text{King})=\frac{2}{46}=\frac{1}{23}$$

Answer: $\boxed{\frac{1}{23}}$

2. Probability of getting a red card

Originally there are 26 red cards. Six red face cards are removed.

Red cards left = 26 − 6 = 20

$$P(\text{Red})=\frac{20}{46}=\frac{10}{23}$$

Answer: $\boxed{\frac{10}{23}}$

3. Probability of getting a spade

No spade is removed, so all 13 spades remain.

$$P(\text{Spade})=\frac{13}{46}$$

Answer: $\boxed{\frac{13}{46}}$

Given:

• Total number of balls = 27
• Probability of getting a red ball = 2/3

Formula

$$\text{Probability} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$$

Let the number of red balls be R.

$$\frac{R}{27}=\frac{2}{3}$$

$$R=\frac{2}{3}\times 27=18$$

Number of white balls:

$$27-18=9$$

Answer

Number of white balls = 9