The probability is the chance of the occurrence of an event. For example, the chance of getting a head while tossing a coin is ½. The value of probability always lies between 0 and 1. In general, the questions in probability are related to rolling a die, tossing a coin, choosing a card from a pack of cards and so on.
The formula of probability is the ratio of the number of favorable outcomes to the total number of outcomes. The terms that define probability are events, outcome, sample space, etc. Students can get the solved problems on probability in the following sections.
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Solved Probability Problems
Problem 1:
A bag contains 8 balls numbered 1 to 8
(i) What is the probability of selecting 1 from the bag?
(ii) What is the probability of selecting an odd number?
(iii) What is the probability of selecting a number less than 4?
Solution:
Total number of bags = 8
Sample sapce S = {1, 2, 3, 4, 5, 6, 7, 8}
(i)
Number of 1 balls = 1
Probability of selecting 1 from the ball = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 1 }{ 8 } \)
(ii)
Number of odd number balls = 5
Probability of selecting odd numbered ball = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 5 }{ 8 } \)
(iii)
Number of balls which are less than 4 = 3
Probability of selecting odd numbered ball = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 3 }{ 8 } \)
Problem 2:
Two coins are tossed simultaneously, find the probability that two heads are obtained.
Solution:
Sample space S = {HT, TH, HH, TT}
Let E be the event of obtaining two heads
So, E = {HH}
Probability of getting two heads = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 1 }{ 4 } \)
Problem 3:
A single card is drawn from a pack of 52 cards. Find the probability of
(a) The card is not king
(b) The card is a red king
(c) The card is king or queen
(d) The card is a diamond
(e) The card is king
(f) The card is either red or ace
(g) The card is black
(h) The card is a 4 or lower
Solution:
The total of playing cards = 52
(a) The card is not king
Number of kings = 4
Probability of getting king = \(\frac { Number of Kings }{ Total number of cards } \)
= \(\frac { 4 }{ 52 } \) = \(\frac { 1 }{ 13 } \)
Probability of the card is not king = 1 – Probability of getting king
= 1 – \(\frac { 1 }{ 13 } \) = \(\frac { 12 }{ 13 } \)
(b) The card is red king
Probability of the card is a red king = \(\frac { Number of Red Kings }{ Total number of cards } \)
= \(\frac { 2 }{ 52 } \) = \(\frac { 1 }{ 26 } \)
(c) The card is king or queen
Number of kings = 4
Number of queens = 4
Total number of kings or queens = 4 + 4 = 8
P(card is a king or queen) = \(\frac { Number of Kings or Queens }{ Total number of cards } \)
= \(\frac { 8 }{ 52 } \) = \(\frac { 2 }{ 13 } \)
(d) The card is a diamond
Number of diamonds = 13
P(card is a diamond) = \(\frac { Number of diamonds }{ Total number of cards } \)
= \(\frac { 13 }{ 52 } \) = \(\frac { 1 }{ 4 } \)
(e) The card is king
Number of kings = 4
P(King) = \(\frac { Number of Kings }{ Total number of cards } \)
= \(\frac { 4 }{ 52 } \) = \(\frac { 1 }{ 13 } \)
(f) The card is either red or ace
Total number of red cards or ace cards = 28
P(card is either red card or ace) = \(\frac { Number of either cae or red cards }{ Total number of cards } \)
= \(\frac { 28 }{ 52 } \) = \(\frac { 7 }{ 13 } \)
(g) The card is black
No. of black cards = 26
P(card is black) = \(\frac { Number of black cards }{ Total number of cards } \)
= \(\frac { 26 }{ 52 } \) = \(\frac { 1 }{ 2 } \)
(h) The card is a 4 or lower
Number of cards is a 4 or lower = 12
P(card is a 4 or lower) = \(\frac { Number of 4 or lower cards }{ Total number of cards } \)
= \(\frac { 12 }{ 52 } \) = \(\frac { 3 }{ 13 } \)
Problem 4:
3 fair coins are tossed. What is the probability of getting at least 2 heads?
Solution:
Sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Total no of outcomes = 8
Let E be the event of getting at least 2 heads = {HHT, HTH, THH, HHH}
Number of favorable outcomes = 4
P(E) = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 4 }{ 8 } \)
= \(\frac { 1 }{ 2 } \)
Therefore, the probability of getting at least 2 heads while tossing 3 coins is \(\frac { 1 }{ 2 } \).
Problem 5:
Two dice are rolled simultaneously. Find the probability of (i) doublet (ii) product of 10 (iii) sum of at least 10?
Solution:
The total number of outcomes = 36
(i) Doublet
The possible doublets are {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} = 6
P(doublets) = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 6 }{ 36 } \)
= \(\frac { 1 }{ 6 } \)
(ii) Product of 10
The possible product of 10 are {(2,5), (5,2)} = 2
P(product of 10) = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 2 }{ 36 } \)
= \(\frac { 1 }{ 18 } \)
(iii) Sum of at least 10
Possibilities of sum of 10 are {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)} = 6
P(sum of at lesat 10) = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 6 }{ 36 } \)
= \(\frac { 1 }{ 6 } \)
Problem 6:
What is the probability of a die showing a number 2 or 6?
Solution:
P(2) is the probability of die showing 2
P(6) is the probability of die showing 6
P(2) = \(\frac { 1 }{ 6 } \), P(6) = \(\frac { 1 }{ 6 } \)
P(2 or 6) = P(2) + P(6)
= \(\frac { 1 }{ 6 } \) + \(\frac { 1 }{ 6 } \)
= \(\frac { 1 }{ 3 } \)