Practice Problems on Probability (Hard)

Probability is an important chapter for the students of Class 9, 10, 11, and 12. The Probability Questions, with their answers included in this article, will help you understand the basic concepts and formulas through a number of solved and unsolved questions. These questions cover concepts like Sample Space, Events, Coin Probability, etc. Solving these problems will improve your understanding and problem-solving skills in probability.

Check - Tricks To Solve Probability Questions

Important Formulas on Probability

Solved Practice Problems on Probability (Hard)

Question 1: A test consists of 10 true/false questions. To pass, a student must answer at least 8 questions correctly. If a student guesses on each question, what is the probability of passing the test?

Solution:

The probability of guessing each question correctly = 1/2.
The probability of getting exactly 8, 9, or 10 questions right can be calculated using the binomial probability formula: P(X=k) = C(n, k) × (p)^k × (1-p)^n-k, where C(n, k) is the combination.
Calculate P(8) + P(9) + P(10) using the formula with n = 10, k = 8, 9, 10, and p = 1/2.
Sum these probabilities to find the total probability of passing.

Question 2: A bag contains 8 blue balls and some pink balls. If the probability of drawing a pink ball is half of the probability of drawing a blue ball then find the number of pink balls in the bag.

Solution:

Let us considered the number of pink balls be n.  

The number of blue balls = 8.

Therefore, the total number of balls present in the bag = n + 8.

Now, the probability of drawing a pink ball, i.e. P(X) = n/n + 8

the probability of drawing blue ball, i.e. P(B) = 8/n + 8

According to the question, the probability of drawing pink

ball is half of the probability of drawing the blue ball

So, P(X) = P(B)/2

n/n+8 = (8/n+8)/2

n/n+8 = (4/n+8)

n = 4.

So, number of pink balls present in the bag is 4.

Question 3: A box contains 5 red, 3 green, and 2 blue balls. Two balls are drawn at random. What is the probability that they are of different colors?

Solution:

Total balls = 5 (red) + 3 (green) + 2 (blue) = 10.

Probability of drawing one red and one green = (5/10) × (3/9) + (3/10) × (5/9).

Probability of drawing one red and one blue = (5/10) × (2/9) + (2/10) × (5/9).

Probability of drawing one green and one blue = (3/10) × (2/9) + (2/10) × (3/9).

Total probability = Sum of the above probabilities.

Question 4: A bag contains 5 red, 6 blue, and 7 green balls. Three balls are drawn without replacement. What is the probability that they are drawn in the order red, blue, and green?

Solution:

Probability of drawing a red ball first = 5/18.

Then, probability of drawing a blue ball = 6/17 (after removing one red ball).

Finally, probability of drawing a green ball = 7/16 (after removing one red and one blue ball).

Total probability = (5/18) × (6/17) × (7/16) ≈ 0.0456 or 4.56%.

Question 5. Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. Find the probability of getting ‘Two tails’.

Solution:

Total number of outcomes = 360

Let us considered the number of times ‘No Tail’ appeared be z

Then, number of times ‘2 Tails’ appeared = 3z

Number of times ‘1 Tail’ appeared = 2z

Now, z + 2z + 3z = 360

6z = 360

z = 60

Hence, the probability of getting ‘two tails’ = (3 x 60)/360 = 1 /2

Question 6: A factory has 3 machines: Machine A, Machine B, and Machine C, producing 50%, 30%, and 20% of the total output, respectively. Machines A, B, and C produce defective items at rates of 1%, 2%, and 3%. If an item is selected randomly and found to be defective, what is the probability it came from Machine B?

Solution:

Let:

• P(B): Probability the item comes from Machine B = 30% = 0.30.
• P(D∣B): Probability of a defective item given it comes from Machine B = 2% = 0.02.
• P(D): Overall probability of a defective item.
• P(B∣D): Probability the defective item came from Machine B (what we need to find).

Total probability of defect: P(D) = P(D∣A) P(A) + P(D∣B) P(B) + P(D∣C) P(C)
P(D) = (0.01) (0.50) + (0.02) (0.30) + (0.03) (0.20) = 0.005 + 0.006 + 0.006 = 0.017

Probability it came from Machine B: P(B∣D) = P(D∣B) P(B) / P(D)
=(0.02) (0.30)/0.017 = 0.006/0.017 ≈ 0.3529

The probability that the defective item came from Machine B is approximately: 0.3529 or 35.29%

Question 7: You have a fair six-sided die and a fair coin. You roll the die and flip the coin simultaneously. If the coin shows heads, you record the number on the die. If the coin shows tails, you roll the die again and record the sum of the two rolls. What is the probability that the recorded number is an even number greater than 6?

Solution:

A fair six-sided die has outcomes {1, 2, 3, 4, 5, 6}.

A fair coin has two equally likely outcomes: heads or tails.

• Coin shows heads: The number is the die roll (1 to 6). No number is both even and greater than 6, so P = 0
• Coin shows tails: The recorded number is X₁ + X₂​ (sum of two rolls, 2 to 12). Outcomes greater than 6 and even are 8, 10, 12.
  • Ways to get 8: (2, 6), (3, 5), (4, 4), (5,3), (6, 2) ⇒ 5 ways.
  • Ways to get 10: (4, 6), (5, 5), (6, 4) ⇒3 ways.
  • Ways to get 12 (6, 6) ⇒1 way.

Probability of recording an even number greater than 6

1. Heads: 0

2. Tails: P(even sum > 6) = 5 + 3 + 1 = 9 outcomes from 36.
= 9/36
=1/4

Combine probabilities:
The coin flip is fair, so: P(heads) = P(tails) = 1/2

Using the law of total probability:
P(even and greater than 6) = P(even and greater than 6 | heads) ⋅ P(heads) + P(even and greater than 6 | tails) ⋅ P(tails)

P(even > 6) = 1/2​⋅0 + 1/2 ​⋅1/4 ​= 1/8 = 0.125

The probability that the recorded number is an even number greater than 6 is 0.125 or 12.5%.

Question 8: A standard deck of 52 cards is shuffled thoroughly. You draw one card, record its suit, and return it to the deck. This process is repeated three times. What is the probability that: All three cards drawn are of the same suit? At least one card drawn is a heart?

Solution:

Probability that all three cards drawn are of the same suit

Each card can belong to one of four suits: {Heart, Spades, Clubs, Diamonds}

Step 1: The probability of drawing a card of any given suit is 1/4​.

Step 2: For all three cards to be the same suit, the first card can be of any suit (1), and the second and third cards must match this suit.

• The probability is: P(same suit for all 3)=1/4 ⋅ 1/4 ⋅ 1/4 = 1/64.
• However, any of the 4 suits could occur, so multiply this result by 4.
  • P(all 3 same suit) = 4 ⋅ 1/64​ = 1/16​.

The probability that all three cards drawn are of the same suitis 1/16 or 6.25%.

Probability that at least one card is a heart

This is the complement of the event that none of the cards drawn is a heart.

Step 1: The probability of not drawing a heart in one draw is:

• P(not a heart) = 1 − P(heart) = 1 - 1/4 =3/4​.

Step 2: The probability that all three cards are not hearts is:

• P(no hearts) = P(not a heart)³ = (3/4​)³ = 27/64​.

Step 3: The probability of at least one heart is:

• P(at least one heart) = 1 − P(no hearts) = 1 − 27/64 = 37/64

Probability at least one card is a heart: 37/64.

Practice More:

Check - Probability Quiz

Practice Problems on Probability (Hard)

Question 1: A quiz consists of 8 multiple-choice questions, each with 4 options. A student guesses all answers. What is the probability of answering exactly 6 questions correctly?

Question 2: A bag contains 10 green and 5 red marbles. If the probability of picking a red marble is doubled, find the total number of marbles in the bag.

Question 3: A box contains 4 black, 5 white, and 6 red balls. Two balls are drawn at random. What is the probability that the balls are of the same color?

Question 4: A bag contains 3 blue, 4 red, and 5 yellow balls. Three balls are drawn one after another without replacement. What is the probability of drawing blue, red, and yellow in that order?

Question 5: Two dice are rolled simultaneously. What is the probability that the sum of the numbers is 9?

Question 6: A company produces 60%, 25%, and 15% of its output from Machines A, B, and C, respectively. The defective rates are 2%, 3%, and 4%. If a randomly selected product is defective, what is the probability it was produced by Machine C?

Question 7: A die is rolled, and a coin is flipped simultaneously. If the die shows a number greater than 3 and the coin shows heads, what is the probability of this outcome?

Question 8: A standard deck of 52 cards is shuffled. Three cards are drawn one by one with replacement. What is the probability that all three are face cards?

Answer Key:

1. 63/16384​ or approximately 0.0034 (or 0.34%)
2. Totalmarbles = 15
3. Probability = 31/105 or approximately 0.2952
4. Probability of 1/22 or approximately 0.0455.
5. Probability of 1/9​ or approximately 0.1111.
6. Probability = 0.2535 or 23.53%
7. Probability = 1/4​ or 0.25%.
8. Probabability of 27/2197​ or approximately 0.0123.