Simple Probability Problems

Probability definition:

If an experiment is performed n is the number of exhaustive cases and m is the number of favourable cases of an event A. Then probability of an event A is defined by

P ( A ) = number of favourable cases / number of exhaustive cases

= n ( A ) / n ( S )

Where n ( A ) = number of elements belonging to A.

n ( S ) = number of elements belonging to S ( Sample space )

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simple probability problems with solutions


Find the probability of getting one red king if we select a card from a pack of 52 cards.

Solution:- There are 2 red kings.

`:.`    Number of possible cases = 2

Number of exhaustive cases =52

`:.` Probability = 2 / 52 = 1 / 26

Find the probability of getting a Head in tossing a coin

Solution:- Number of exhaustive cases = 2

Number of possible cases = 1

`:.` Probability = 1 / 2

Find the probability of getting a sum 8 if two dices are thrown .

Solution:-  The total number of outcomes = 62 = 36

Favourable outcomes = { ( 2, 6 ), ( 6, 2 ), ( 3, 5 ), ( 5, 3 ), ( 4, 4 ) }

Number of favourable outcomes = 5

`:.` Probability of getting sum 8 = 5 / 36

What is the probability that a leap year selected at random will contain 53 Sundays.

Solution:- A leap year contains 366 days

366 days = 52 weeks + 2 days

The possibility is Sunday-Monday, Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday, Thursday-Friday, Friday-Saturday, Saturday-Sunday

P ( Sunday ) = 2 / 7

Find the probability of getting one head in tossing two coins

Solution:- Let A be the event of getting one head

A = [ HT, TH ]

S = [ HH, HT, TH, TT ]

Number of elements in S = 4

number of elements in A = 2

`:.` Probability of an event A = 2 / 4 = 1/ 2.

What is the probability of drawing an ace from a well shuffled deck of 52 playing cards?

Solution:- Number of exhaustive cases = 52C1 =52

Number of possible cases = 4C1 = 4     ( since there are 4 aces )

`:.` Required probability = 4 / 52 = 1/ 13.

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