Maths Class 10 Chapter 15: Probability

# Maths Class 10 Chapter 15: Probability

## Probability Formulas

1.      The probability of an event which is based on the result of an actual experiment is called an experimental probability. For example, in an experiment of tossing a coin 1000 times, the outcomes are as follows: Head : 485 and Tail : 515. The probability of getting head is 0.485 and the probability of getting tail is 0.515. These probabilities are called experimental probabilities.

2.      If all the outcomes of an event have equal chances to occur, then the outcomes are called the equally likely outcomes.

3.      If the probability is calculated using a formula, then this probability is called a theoretical probability. The theoretical probability of an event E, written as P(E), is defined as follows:

where we assume that the outcomes of the experiment are equally likely.

4.      The probability of an event E is a number P(E) such that     0 P(E) 1

5.      An event having only one outcome is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is 1.

6.      For any event E,

## Problems on Probability

Example 1: Three coins are tossed simultaneously. Find the probability of the event of getting
c. one head and two tails         d. three tails

Solution:  When three coins are tossed, the possible outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
Here, total number of outcomes = 8
a. Probability of getting three heads = 1/8                  (Favorable outcome is HHH)
b. Probability of getting two heads and one tail = 3/8 (Favorable outcomes are HHT, HTH, THH)
c. Probability of getting one head and two tails = 3/8 (Favorable outcomes are TTH, THT, HTT)
d. Probability of getting three tails = 1/8                     (Favorable outcome is TTT)

Example 2: A fair dice is rolled. Find the probability of getting
a. a 6
b. a number less than 3
c. an odd number

Solution:
a. In rolling a dice, there are six equally likely outcomes,
which are 1, 2, 3, 4, 5 and 6.
The event of getting a 6 consists of the one outcome ‘6’.
The probability of getting a 6, P(getting a 6) = 1/6 .

b. The favorable outcomes of the event of getting a number less than 3 are 1 and 2.
The probability of getting a number less than 3,
P(getting a number less than 3) = 2/6 = 1/3

c. There are three favorable outcomes for the event of getting an odd number. They
are 1, 3 and 5.
The probability of getting an odd number,
P(getting an odd number) = 3/6 = 1/2
Example 3: There are 5 red, 6 green and 7 blue balls in a bag. One ball is drawn at random. What is the probability that the ball drawn is a
a. red ball,                      b. green ball,                 c. not a green ball?

Solution:  Total number of balls in the bag = 5 + 6 + 7 = 18
a. P(red) = n(E)/n(S) = 5/18
b. P(green) = n(E)/n(S) = 6/18 = 1/3
c. Out of 18 balls, 12 are not green.
Thus, P(not green) = 12/18 =  2/3

Example 4: In a box, there are 20 cards bearing numbers from 1 to 20. A card is taken out of the box at random. What is the probability that  the card taken out is
a. a multiple of 3        b. a prime number         c. a multiple of 7        d. an odd number

Solution: Total number of cards in the box = 20
a. Multiples of 3 are 3, 6, 9, 12, 15, 18, i.e. 6 multiples in all
Thus, P (getting multiple of 3) = 6/20 = 3/10
b. Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, i.e. 8 in all
Thus, P (getting prime number) = 8/20 = 2/5
c. Multiples of 7 are 7, 14, i.e. 2 in all
Thus, P (getting multiple of 7) = 2/20 = 1/10
d. Odd numbers are 10 in numbers
Thus, P (getting odd numbers) = 10/20 = 1/2

Example 5: In a pack of cards, there are 52 cards. Find the probability of getting
a.  a red card        b. a spade card

Solution:  a. In a pack of cards, there are 26 red cards.
Number of favorable outcomes = 26
P (Red card) = 26/52 = 1/2

b.  In a pack of cards, there are 13 spade cards
Number of favorable outcomes = 13
P (Spade card) = 13/52 = 1/3

Example 6: Write the sample space when two coins are tossed. Find the probability of getting at least one head.

Solution:  Sample space = {HH, HT, TH, TT}, where H = head and T = tail
Event = {HH, HT, TH}
Thus, n(E) = 3 and n(S) = 4
P (getting head) = n(E)/n(S) = 3/4

Example 7: A card is drawn randomly from a deck of 52 playing cards. Find the probability that the card is
a. black,              b. an ace,                 c. a red queen.

Solution:  a. In a deck of 52 playing cards, there are 26 black cards, i.e. 13 spade and 13 club.
Thus, n(E) = 26 and n(S) = 52
P (getting black card) = n(E)/n(S) = 26/52 = ½

b. There are 4 aces in the deck of 52 playing cards.
Thus, n(E) = 4 and n(S) = 52
P (getting an ace) = n(E)/n(S) = 4/52 = 1/13

c. There are 2 red queens in the deck of 52 playing cards.
Thus, n(E) = 2 and n(S) = 52

P (getting a red queen) = n(E)/n(S) = 2/52 = 1/26