Chapter 14: Probability

A survey was conducted among 50 students of Class 10 to determine their blood groups. The data obtained is recorded in the table below:

1. Find the probability that a student selected at random has blood group AB.
   (A) 0.16

   (B) 0.24

   (C) 0.30

   (D) 0.08
   Show Solution
   Solution: (A) 0.16
   Step 1: Total number of students = 50.
   Step 2: Number of students with blood group AB = 8.
   Step 3: Probability $P(AB) = \frac{8}{50} = \frac{4}{25} = 0.16$.
2. Find the probability that a student selected at random has blood group O or B.
   (A) 0.3

   (B) 0.6

   (C) 0.5

   (D) 0.4
   Show Solution
   Solution: (B) 0.6
   Step 1: Number of students with blood group O = 15.
   Step 2: Number of students with blood group B = 15.
   Step 3: Total favorable outcomes = $15 + 15 = 30$.
   Step 4: Probability $P(O \text{ or } B) = \frac{30}{50} = 0.6$.

All kings, queens, and aces are removed from a standard pack of 52 playing cards. The remaining cards are well shuffled, and then one card is drawn at random.

1. Find the probability that the drawn card is a black face card.
   (A) 1/20

   (B) 1/10

   (C) 1/26

   (D) 3/26
   Show Solution
   Solution: (A) 1/20
   Step 1: Cards removed: 4 Kings + 4 Queens + 4 Aces = 12 cards.
   Step 2: Remaining cards = $52 - 12 = 40$.
   Step 3: Face cards are King, Queen, Jack. Since Kings and Queens are removed, only Jacks remain.
   Step 4: Black Jacks are Jack of Spades and Jack of Clubs (2 cards).
   Step 5: Probability = $\frac{2}{40} = \frac{1}{20}$.
2. Find the probability that the drawn card is a red card.
   (A) 1/2

   (B) 13/20

   (C) 1/4

   (D) 2/5
   Show Solution
   Solution: (A) 1/2
   Step 1: Total red cards in a deck = 26.
   Step 2: Red cards removed: 2 Kings + 2 Queens + 2 Aces = 6 cards.
   Step 3: Remaining red cards = $26 - 6 = 20$.
   Step 4: Probability = $\frac{20}{40} = \frac{1}{2}$.

Two dice are thrown at the same time. The numbers appearing on the top faces of the dice are noted.

1. Find the probability that the sum of the two numbers is 8.
   (A) 1/6

   (B) 5/36

   (C) 1/9

   (D) 1/12
   Show Solution
   Solution: (B) 5/36
   Step 1: Total possible outcomes = $6 \times 6 = 36$.
   Step 2: Favorable outcomes for sum 8: $(2,6), (3,5), (4,4), (5,3), (6,2)$.
   Step 3: Number of favorable outcomes = 5.
   Step 4: Probability = $\frac{5}{36}$.
2. Find the probability that the product of the two numbers is a prime number.
   (A) 1/6

   (B) 5/36

   (C) 7/36

   (D) 2/9
   Show Solution
   Solution: (A) 1/6
   Step 1: A product is prime only if one number is 1 and the other is a prime number.
   Step 2: Prime numbers on a die: 2, 3, 5.
   Step 3: Favorable outcomes: $(1,2), (2,1), (1,3), (3,1), (1,5), (5,1)$.
   Step 4: Number of favorable outcomes = 6.
   Step 5: Probability = $\frac{6}{36} = \frac{1}{6}$.