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
a. three heads b. two
heads and one tail
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
Thanks for this beneficial article.
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