Chapter 19 Probability

(i) The probability of an impossible event is zero.

(ii) The probability of a sure event is one.

(iii) For any event E, P(E) + P(not E)= one .

(iv) The probability of a possible but not a sure event lies between zero and one.

(v) The sum of probabilities of all the outcomes of an experiment is one.

In a throw of a dice, all possible outcomes are 1, 2, 3, 4, 5, 6

Total number of possible outcomes = 6

(i)Let E be event of getting even number

Then, the favorable outcomes are 2, 4, 6

Number of favorable outcomes= 3

∴ P(getting a even number)= P(E) =  

(ii)Let R be the number less than 5

Then, the favorable outcomes are 1, 2, 3, 4

Number of favorable outcomes = 4

∴ P(getting a number less than 5)= P(R) =  

(iii)Let M be the event of getting a number greater than 2

Then, the favorable outcomes are 3, 4, 5, 6

Number of favorableoutcomes = 4

∴ P(getting a number greater than 2)= P(M)  

(iv)Let N be the number lying between 3 and 6

Then the favorable outcomes are 4, 5

Number of favorable outcomes = 2

∴ P(getting a number 3 and 6)= P(N) =

(v)Let G be event of getting a number other than 3

Then the favorable outcomes are 1, 2, 4, 5, 6

Number of favorable outcomes = 5

∴ P(getting a number other than 5)=P(G) =  

(vi)Let T be event of getting a number 5

Then the favorable outcome is 5

Number of favorable outcomes = 1

∴ P(getting a number 5)=P(T)  

Outcomes are A, B, C, A, A, B

Total number of outcomes = 6

(i) Let E₁ be the event of getting a letter 'A'.

Then, the number of favourable outcomes = 3

(ii) Let E₂ be the event of getting a letter 'B'.

Then, the number of favourable outcomes = 2

Number of defective pens = 12

Number of good pens = 132

Then, total number of pens = 12 + 132 = 144

(i) Let E₁ be the event of getting a good pen.

Then, the number of favourable outcomes = 132

(ii) Let E₂ be the event of getting a defective pen.

Then, the number of favourable outcomes = 12

Total number of tickets sold = 250

Number of prizes = 5

Let E be the event getting a prize

Number of favorable outcomes = 5

P(getting a prize) =

Since the arrow can come to rest at any one of the numbers 1, 2, 3, 4, ……, 12.

∴ Total number of outcomes = 12

(i) Let E₁ be the event of pointing at number 6.

Then, the number of favourable outcomes = 1

∴ P(E₁) =  

(ii) Let E₂ be the event of pointing at an even number.

Then, favourable outcomes are 2, 4, 6, 8, 10, 12.

So, number of favourable outcomes = 6

∴ P(E₂) =  

(iii) Let E₃ be the event of pointing at a prime number.

Then, favourable outcomes are 2, 3, 5, 7, 11

So, number of favourable outcomes = 5

∴ P(E₃) =  

(iv) Let E₄ be the event of pointing at a number which is a multiple of 5.

Then, favourable outcomes are 5, 10

So, number of favourable outcomes = 2

∴ P(E₄) =  

(v) Let E₅ be the event of pointing at a number which is a factor of 8.

Then, favourable outcomes are 1, 2, 4, 8

So, number of favourable outcomes = 4

∴ P(E₅) =  

Outcomes are 1, 2, 3, ....., 17

Total number of outcomes = 17

(i) Let E₁ be the event of getting an odd numbers.

Then, favourable outcomes are 1, 3, 5, 7, 9, 11, 13, 15, 17.

So, number of favourable outcomes = 9

(ii) Let E₂ be the event of getting a number divisible by 5.

Then, favourable outcomes are 5, 10, 15.

So, number of favourable outcomes = 3

Number of white balls = 5

Number of red balls = 7

Number of black balls = 4

Number of blue balls = 2

Then, total number of balls = 5 + 7 + 4 + 2 = 18

(i) Let E₁ be the event of drawing a white or blue ball.

Then, the number of favourable outcomes = 5 + 2 = 7

∴ P(E₁) =  

(ii) Let E₂ be the event of drawing neither white nor black.

Then, the number of balls which are neither white nor black = 7 + 2 = 9

∴ P(E₂)  

Total number of marbles = 24

Given,

⇒ Number of green marbles = 16

⇒ Number of blue marbles = 24 - 16 = 8

Total number of marbles = 54

Given,

⇒ Number of blue marbles = 18

Also,

⇒ Number of green marbles = 24

Therefore, number of white marbles in the jar = 54 - 18 - 24 = 12

Let the number of black balls in the bag = x

Number of white balls in the bag = 15

So, total number of balls in the bag = x + 15

Given,

P(drawing a black ball) = 3 × P(drawing a white ball)

Hence, the number of black balls in the bag is 45.

Number of white balls in the bag = x

Number of black balls in the bag = 2x

Number of red balls in the bag = 3x

So, total number of balls in the bag = x + 2x + 3x = 6x

(i) Let E₁ be the event of not drawing a red ball.

Then, the number of balls which are not red = 6x - 3x = 3x

∴ P(E₁)  

(ii) Let E₂ be the event of drawing a white ball.

Then, the number of favourable outcomes = x

∴ P(E₂)  

Total number of cards = 20

(i) Let E₁ be the event of drawing a prime number card.

Then, the favourable outcomes are 2, 3, 5, 7, 11, 13, 17, 19

So, number of favourable outcomes = 8

∴ P(E₁)  

(ii) Let E₂ be the event of drawing a composite number card.

Then, the favourable outcomes are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20

So, number of favourable outcomes = 11

∴ P(E₂)  

(iii) Let E₃ be the event of drawing a number card that is divisible by 3.

Then, the favourable outcomes are 3, 6, 9, 12, 15, 18

So, number of favourable outcomes = 6

∴ P(E₃)  

Total number of tickets = 100

(i)Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100

Total number of even number = 50

P(getting a even number) =

(ii)Numbers less than 16 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

Total number of numbers less than 16 is 14

P(getting a number less than 16) =

(iii)Numbers which are perfect square are 4, 9, 16, 25, 36, 49, 64, 81, 100

Total number of perfect squares = 9

P(getting a perfect square) =

(iv)Prime numbers less than 40 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37

Total number of prime numbers =12

P(getting a prime number less 40) =

When two dice are thrown, the outcomes are as follows:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).

So, total number of outcomes = 36

(i) Let E₁ be the event of obtaining numbers giving even sum.

Then, the favourable outcomes are as follows:

(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6),

(3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6),

(5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6).

So, number of favourable outcomes = 18

∴ P(E₁)  

(ii) Let E₂ be the event of obtaining numbers giving even product.

Then, the favourable outcomes are as follows:

(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).

So, number of favourable outcomes = 27

∴ P(E₂)  

When two dice are thrown, the outcomes are as follows:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).

So, total number of outcomes = 36

(i) Let E₁ be the event of obtaining numbers having sum less than 7.

Then, the favourable outcomes are as follows:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4),

(3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1),

So, number of favourable outcomes = 15

∴ P(E₁)  

(ii) Let E₂ be the event of obtaining numbers having product less than 16.

Then, the favourable outcomes are as follows:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1),

(4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2).

So, number of favourable outcomes = 25

∴ P(E₂)  

(iii) Let E₃ be the event of obtaining a doublet of odd numbers.

Then, the favourable outcomes are as follows:

(1, 1), (3, 3), (5, 5)

So, number of favourable outcomes = 3

∴ P(E₂)  

When two dice are thrown by Peter, the outcomes are as follows:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).

So, total number of outcomes = 36

Let E₁ be the event of obtaining numbers having product as 25.

Then, the favourable outcome is (5, 5).

So, number of favourable outcomes = 1

∴ P(E₁)  

When a die is thrown by Rina, the outcomes are as follows:

1, 2, 3, 4, 5, 6

When these numbers are squared, we have

1, 4, 9, 16, 25, 36

⇒ Total number of outcomes = 6

Let E₂ be the event of obtaining a number 25.

Then, the favourable outcome is 25.

So, number of favourable outcomes = 1

∴ P(E₂)  

Since  

⇒ P(E₂) > P(E₁)

Hence, we conclude that Rina has better chance than Peter to get the number 25.

When a pair of distinct dice is thrown, the outcomes are as follows:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).

So, total number of outcomes = 36

Let E be the event of getting the sum of numbers either less than 3 or more than 11.

Then, the favourable outcome are (1, 1) and (6, 6).

So, number of favourable outcomes = 2

∴ P(E)  

Total number of apples = 900

Let E be the event of selecting a rotten apple from a heap.

∴ P(E) = 0.18

Hence, the number of rotten apples in the heap is 162.

An ordinary year has 365 days which means 52 complete weeks and one day.

That means, there are 52 Sundays in 52 weeks.

This 'one day' can be either Sunday, Monday, Tuesday, Wednesday, Thursday, Friday or Saturday.

So, we have

Total number of outcomes = 7

Number of favourable outcomes = 1

∴ Probability that an ordinary year has 53 Sundays  

When a coin is tossed three times, the possible outcomes are as follows:

HHH, HTH, HHT, THH, THT, TTH, HTT, TTT

So, we have

Total number of outcomes = 8

Getting the same result in all the tosses is a success.

And, not getting the same result in all the tosses is losing the game.

So, favourable outcomes are HTH, HHT, THH, THT, TTH, HTT.

⇒ Number of favourable outcomes = 6

∴ Probability of losing the game  

Total number of cards in a pack after losing cards of king, jack and 10 of spades = 52 - 3 = 49

(i) Let E₁ be the event of drawing a red card.

Number of red cards in a pack = 26

∴ P(E₁)  

(ii) Let E₂ be the event of drawing a black jack.

Number of black jack cards in a pack after losing a jack of spades = 2 - 1 = 1

∴ P(E₂)  

(iii) Let E₃ be the event of drawing a red king.

Number of red king cards in a pack = 2

∴ P(E₃)  

(iv) Let E₄ be the event of drawing a 10 of hearts card.

Number of 10 of heart cards in a pack = 1

∴ P(E₄)  

There are 26 red cards containing a 2 queensand 2 more black queens are there in a pack of cards

P(getting a red card or a queen)

P(getting neither a red card nor a queen) =

Total number of cards = 52

(i)There are 13 cards of spade (including 1 ace) and 3 more ace cards are there in a pack of cards

P(getting a card of spades or an ace) =

(ii)There are 2 red kings in a pack of cards

P(getting a red king) =

(iii)There are 4 kings and 4 queens in a pack of cards

P(getting either a king or a queen) =

(iv)P(getting neither a king nor a queen) =

Total numbers = 10

Sum of given numbers = 3 + 4 + 4 + 4 + 5 + 5 + 6 + 6 + 6 + 7 = 50

∴ Mean  

Let E be the event of selecting a number 5.

So, number of favourable outcomes = 2

∴ P(E)  

Let E be the event that it will rain tomorrow.

∴ P(E) = 0.85

⇒ Probability that it will not rain tomorrow = P(not E) = 1 - P(E) = 1 - 0.85 = 0.15

A leap year has 366 days which means 52 complete weeks and 2 days.

That means, there are 52 Sundays and 52 Mondays in 52 weeks.

These '2 days' can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)

So, we have

Total number of outcomes = 7

Number of favourable outcomes (Sunday, Monday) = 1

∴ Probability that a leap year has 53 Sundays and 53 Mondays