Chapter 31 Probability Distribution

(i) Given :

To find : mean (𝓊), variance (σ²) and standard deviation (σ)

Formula used :

Mean = E(X) =

Variance = E(X²) -

Standard deviation =

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃) + x₄P(x₄)

Mean = E(X) = 0() + 1() +2() +3() = 0 + + + = = =

Mean = E(X) = = 1.2

= = 1.44

E(X²) = = P(x₁) + P(x₂) + P(x₃) + P(x₄)

E(X²) = () + () + () + () = 0 + + + = =

E(X²) = 2

Variance = E(X²) - = 2 – 1.44 = 0.56

Variance = E(X²) - = 0.56

Standard deviation = = = 0.74

Mean = 1.2

Variance = 0.56

Standard deviation = 0.74

(ii) Given :

To find : mean (𝓊), variance (σ²) and standard deviation (σ)

Formula used :

Mean = E(X) =

Variance = E(X²) -

Standard deviation =

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃) + x₄P(x₄)

Mean = E(X) = 1(0.4) + 2(0.3) +3(0.2) +4(0.1) = 0.4 + 0.6 + 0.6 + 0.4 = 2

Mean = E(X) = 2

= = 4

E(X²) = = P(x₁) + P(x₂) + P(x₃) + P(x₄)

E(X²) = (0.4) + (0.3) + (0.2) + (0.1) = 0.4 + 1.2 + 1.8 + 1.6 = 5

E(X²) = 5

Variance = E(X²) - = 5 – 4 = 1

Variance = E(X²) - = 1

Standard deviation = = = 1

Mean = 2

Variance = 1

Standard deviation = 1

(iii) Given :

To find : mean (𝓊), variance (σ²) and standard deviation (σ)

Formula used :

Mean = E(X) =

Variance = E(X²) -

Standard deviation =

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃) + x₄P(x₄)

Mean = E(X) = -3(0.2) + (-1)(0.4) + 0(0.3) + 2(0.1)= -0.6 - 0.4 + 0 + 0.2 = -0.8

Mean = E(X) = -0.8

= = 0.64

E(X²) = = P(x₁) + P(x₂) + P(x₃) + P(x₄)

E(X²)= (0.2) + (0.4) + (0.3) + (0.1) = 1.8 + 0.4 + 0+ 0.4 = 2.6

E(X²) = 2.6

Variance = E(X²) - = 2.6 – 0.64 = 1.96

Variance = E(X²) - = 1.96

Standard deviation = = = 1.4

Mean = -0.8

Variance = 1.96

Standard deviation = 1.4

(iv) Given :

To find : mean (𝓊), variance (σ²) and standard deviation (σ)

Formula used :

Mean = E(X) =

Variance = E(X²) -

Standard deviation =

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃) + x₄P(x₄) + x₅P(x₅)

Mean = E(X) = -2(0.1) + (-1)(0.2) + 0(0.4) + 1(0.2) + 2(0.1)

Mean = E(X) = -0.2 - 0.2 + 0 + 0.2 + 0.2 = 0

Mean = E(X) = 0

= = 0

E(X²) = = P(x₁) + P(x₂) + P(x₃) + P(x₄) + P(x₅)

E(X²) = (0.1) + (0.2) + (0.4) + (0.2) + (0.1)

E(X²) = 0.4 + 0.2 + 0 + 0.2 +0.4 = 1.2

E(X²) = 1.2

Variance = E(X²) - = 1.2 – 0 = 1.2

Variance = E(X²) - = 1.2

Standard deviation = = = 1.095

Mean = 0

Variance = 1.2

Standard deviation = 1.095

Given : Two coins are tossed simultaneously

To find : mean (𝓊), variance (σ²)

Formula used :

Mean = E(X) =

Variance = E(X²) -

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃)

When two coins are tossed simultaneously,

Total possible outcomes = TT , TH , HT , HH where H denotes head and T denotes tail.

P(0) = (zero heads = 1 [TT] )

P(1) = (one heads = 2 [HT , TH] )

P(2) = (two heads = 1 [HH] )

The probability distribution table is as follows,

Mean = E(X) = 0() + 1() +2() = 0 + + = = 1

Mean = E(X) = 1

= = 1

E(X²) = = P(x₁) + P(x₂) + P(x₃)

E(X²) = () + () + () = 0 + + = = = 1.5

E(X²) = 1.5

Variance = E(X²) - = 1.5 – 1 = 0.5

Variance = E(X²) - = 0.5

Mean = 1

Variance = 0.5

Given : Three coins are tossed simultaneously

To find : mean (𝓊) and variance (σ²)

Formula used :

Mean = E(X) =

Variance = E(X²) -

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃)

When three coins are tossed simultaneously,

Total possible outcomes = TTT , TTH , THT , HTT , THH , HTH , HHT , HHH where H denotes head and T denotes tail.

P(0) = (zero tails = 1 [HHH] )

P(1) = (one tail = 3 [HTH , THH , HHT ] )

P(2) = (two tail = 3 [HTT , THT , TTH ] )

P(3) = (three tails = 1 [TTT] )

The probability distribution table is as follows,

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃) + x₄P(x₄)

Mean = E(X) = 0() + 1() +2() +3() = 0 + + + = = =

Mean = E(X) = = 1.5

= = 2.25

E(X²) = = P(x₁) + P(x₂) + P(x₃) + P(x₄)

E(X²) = () + () + () + () = 0 + + + = = = 3

E(X²) = 3

Variance = E(X²) - = 3 – 2.25 = 0.75

Variance = E(X²) - = 0.75

Mean = 1.5

Variance = 0.75

Given : A die is tossed twice and ‘Getting an odd number on a toss’ is considered a success.

To find : probability distribution of the number of successes and mean (𝓊) and variance (σ²)

Formula used :

Mean = E(X) =

Variance = E(X²) -

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃)

When a die is tossed twice,

Total possible outcomes =

{(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)}

‘Getting an odd number on a toss’ is considered a success.

P(0) = = (zero odd numbers = 9 )

P(1) = = (one odd number = 18 )

P(2) = = (two odd numbers = 9 )

The probability distribution table is as follows,

Mean = E(X) = 0() + 1() +2() = 0 + + = = 1

Mean = E(X) = 1

= = 1

E(X²) = = P(x₁) + P(x₂) + P(x₃)

E(X²) = () + () + () = 0 + + = = = 1.5

E(X²) = 1.5

Variance = E(X²) - = 1.5 – 1 = 0.5

Variance = E(X²) - = 0.5

The probability distribution table is as follows,

Mean = 1

Variance = 0.5

Given : A die is tossed twice and ‘Getting a number greater than 4 ’ is considered a success.

To find : probability distribution of the number of successes and mean (𝓊) and variance (σ²)

Formula used :

Mean = E(X) =

Variance = E(X²) -

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃)

When a die is tossed twice,

Total possible outcomes =

{(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)}

‘Getting a number greater than 4’ is considered a success.

P(0) = = (zero numbers greater than 4 = 16 )

P(1) = = (one number greater than 4= 16 )

P(2) = = (two numbers greater than 4= 4 )

The probability distribution table is as follows,

Mean = E(X) = 0() + 1() +2() = 0 + + = = =

Mean = E(X) =

= =

E(X²) = = P(x₁) + P(x₂) + P(x₃)

E(X²) = () + () + () = 0 + + =

E(X²) =

Variance = E(X²) - = – =

Variance = E(X²) - =

The probability distribution table is as follows,

Mean =

Variance =

Given : A die is tossed twice and ‘Getting a number greater than 4 ’ is considered a success.

To find : probability distribution of the number of successes and mean (𝓊) and variance (σ²)

Formula used :

Mean = E(X) =

Variance = E(X²) -

When a die is tossed 4 times,

Total possible outcomes = 6² = 36

Getting a doublet is considered as a success

The possible doublets are (1,1) , (2,2) , (3,3) , (4,4) , (5,5) , (6,6)

Let p be the probability of success,

p = =

q = 1 – p = 1 - =

q =

since the die is thrown 4 times, n = 4

x can take the values of 1,2,3,4

P(x) = ⁿC_x

P(0) = ⁴C₀ =

P(1) = ⁴C₁ = =

P(2) = ⁴C₂ = =

P(3) = ⁴C₃ = =

P(4) = ⁴C₄ =

The probability distribution table is as follows,

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃) + x₄P(x₄) + x₅P(x₅)

Mean = E(X) = 0() + 1() + 2() + 3() + 4()

Mean = E(X) = 0 + + + + = = =

Mean = E(X) =

= =

E(X²) = = P(x₁) + P(x₂) + P(x₃) + P(x₄) + P(x₅)

E(X²) = () + () + () + () + ()

E(X²) = 0 + + + + = =

E(X²) = 1

Variance = E(X²) - = 1 – =

Variance = E(X²) - =

The probability distribution table is as follows,

Mean =

Variance =

Given : A coin is tossed 4 times

To find : probability distribution of X and mean (𝓊) and variance (σ²)

Formula used :

Mean = E(X) =

Variance = E(X²) -

A coin is tossed 4 times,

Total possible outcomes = 2⁴ = 16

X denotes the number of heads

Let p be the probability of getting a head,

p =

q = 1 – p = 1 - =

q =

since the coin is tossed 4 times, n = 4

X can take the values of 1,2,3,4

P(x) = ⁿC_x

P(0) = ⁴C₀ =

P(1) = ⁴C₁ = =

P(2) = ⁴C₂ = =

P(3) = ⁴C₃ = =

P(4) = ⁴C₄ =

The probability distribution table is as follows,

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃) + x₄P(x₄) + x₅P(x₅)

Mean = E(X) = 0() + 1() + 2() + 3() + 4()

Mean = E(X) = 0 + + + + = = = 2

Mean = E(X) = 2

= = 4

E(X²) = = P(x₁) + P(x₂) + P(x₃) + P(x₄) + P(x₅)

E(X²) = () + () + () + () + ()

E(X²) = 0 + + + + = = = 5

E(X²) = 5

Variance = E(X²) - = 5 – 4 = 1

Variance = E(X²) - = 1

The probability distribution table is as follows,

Mean = 2

Variance = 1

Given : Let X denote the number of times ‘a total of 9’ appears in two throws of a pair of dice

To find : probability distribution of X ,mean (𝓊) and variance (σ²) and standard deviation

Formula used :

Mean = E(X) =

Variance = E(X²) -

Standard deviation =

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃)

When a die is tossed twice,

Total possible outcomes =

{(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)}

Let X denote the number of times ‘a total of 9’ appears in two throws of a pair of dice

p = =

q = 1 - =

Two dice are tossed twice, hence n = 2

P(0) = ²C₀ =

P(1) = ²C₁ =

P(2) = ²C₂ =

The probability distribution table is as follows,

Mean = E(X) = 0() + 1() +2() = 0 + + = = =

Mean = E(X) =

= =

E(X²) = = P(x₁) + P(x₂) + P(x₃)

E(X²) = () + () + () = 0 + + =

E(X²) =

Variance = E(X²) - = – =

Variance = E(X²) - =

Standard deviation = = =

The probability distribution table is as follows,

Mean =

Variance =

Standard deviation =

Given : There are 5 cards, numbers 1 to 5, one number on each card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn.

To find : mean (𝓊) and variance (σ²) of X

Formula used :

Mean = E(X) =

Variance = E(X²) -

There are 5 cards, numbers 1 to 5, one number on each card. Two cards are drawn at random without replacement.

X denote the sum of the numbers on two cards drawn

The minimum value of X will be 3 as the two cards drawn are 1 and 2

The maximum value of X will be 9 as the two cards drawn are 4 and 5

For X = 3 the two cards can be (1,2) and (2,1)

For X = 4 the two cards can be (1,3) and (3,1)

For X = 5 the two cards can be (1,4) , (4,1) , (2,3) and (3,2)

For X = 6 the two cards can be (1,5) , (5,1) , (2,4) and (4,2)

For X = 7 the two cards can be (3,4) , (4,3) , (2,5) and (5,2)

For X = 8 the two cards can be (5,3) and (3,5)

For X = 9 the two cards can be (4,5) and (4,5)

Total outcomes = 20

P(3) = =

P(4) = =

P(5) = =

P(6) = =

P(7) = =

P(8) = =

P(9) = =

The probability distribution table is as follows,

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃) + x₄P(x₄) + x₅P(x₅) + x₆P(x₆) + x₇P(x₇)

Mean = E(X) = 3() + 4() + 5() + 6() + 7() + 8() + 9()

Mean = E(X) = + + + + + + = = = 6

Mean = E(X) = 6

= = 36

E(X²) = = P(x₁) + P(x₂) + P(x₃) + P(x₄) + P(x₅) + P(x₆) + P(x₇)

E(X²) = () + () + () + () + () + () + ()

E(X²) = + + + + + + = = = 39

E(X²) = 39

Variance = E(X²) - = 39 – 36 = 3

Variance = E(X²) - = 3

Mean = 6

Variance = 3

Given : Two cards are drawn from a well-shuffled pack of 52 cards.

To find : probability distribution of the number of kings and variance (σ²)

Formula used :

Mean = E(X) =

Variance = E(X²) -

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃)

Two cards are drawn from a well-shuffled pack of 52 cards.

Let X denote the number of kings in the two cards

There are 4 king cards present in a pack of well-shuffled pack of 52 cards.

P(0) = = =

P(1) = = =

P(2) = = =

The probability distribution table is as follows,

Mean = E(X) = 0() + 1() +2() = 0 + + = =

Mean = E(X) =

= =

E(X²) = = P(x₁) + P(x₂) + P(x₃)

E(X²) = () + () + () = 0 + + =

E(X²) =

Variance = E(X²) - = – = = =

Variance = E(X²) - =

The probability distribution table is as follows,

Variance =

Given : A box contains 16 bulbs, out of which 4 bulbs are defective. Three bulbs are drawn at random

To find : mean (𝓊) and variance (σ²)

Formula used :

Mean = E(X) =

Variance = E(X²) -

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃)

A box contains 16 bulbs, out of which 4 bulbs are defective. Three bulbs are drawn at random

Let X denote the number of defective bulbs drawn

There are 4 defective bulbs present in 16 bulbs

P(0) = = =

P(1) = = =

P(2) = = =

P(3) = = =

The probability distribution table is as follows,

Mean = E(X) = 0() + 1() + 2() + 3() = 0 + + + =

Mean = E(X) = =

= =

E(X²) = = P(x₁) + P(x₂) + P(x₃)

E(X²) = () + () + () + () = 0 + + + =

E(X²) =

Variance = E(X²) - = – = = =

Variance = E(X²) - =

Mean = E(X) =

Variance =

Given : 20% of the bulbs produced by a machine are defective.

To find probability distribution of a number of defective bulbs in a sample of 4 bulbs chosen at random.

Formula used :

The probability distribution table is given by ,

Where P(x) = ⁿC_x

Here p is the probability of getting a defective bulb.

q = 1 – p

Let the total number of bulbs produced by a machine be x

20% of the bulbs produced by a machine are defective.

Number of defective bulbs produced by a machine = =

X denotes the number of defective bulbs in a sample of 4 bulbs chosen at random.

Let p be the probability of getting a defective bulb,

p = =

p =

q = 1 – p = 1 - =

q =

since 4 bulbs are chosen at random, n = 4

X can take the values of 0,1,2,3,4

P(x) = ⁿC_x

P(0) = ⁴C₀ =

P(1) = ⁴C₁ =

P(2) = ⁴C₂ =

P(3) = ⁴C₃ =

P(4) = ⁴C₄ =

The probability distribution table is as follows,

Given : Four bad eggs are mixed with 10 good ones. Three eggs are drawn one by one without replacement.

To find : mean (𝓊) and variance (σ²)

Formula used :

Mean = E(X) =

Variance = E(X²) -

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃)

Four bad eggs are mixed with 10 good ones. Three eggs are drawn one by one without replacement.

Let X denote the number of bad eggs drawn

There are 4 bad eggs present in 14 eggs

P(0) = = =

P(1) = = =

P(2) = = =

P(3) = = =

The probability distribution table is as follows,

Mean = E(X) = 0() + 1() + 2() + 3() = 0 + + + =

Mean = E(X) = =

= =

E(X²) = = P(x₁) + P(x₂) + P(x₃)

E(X²) = () + () + () + () = 0 + + + =

E(X²) =

Variance = E(X²) - = – = =

Variance = E(X²) - =

Mean = E(X) =

Variance =

Given : Four rotten oranges are mixed with 16 good ones. Three oranges are drawn one by one without replacement.

To find : mean (𝓊) and variance (σ²)

Formula used :

Mean = E(X) =

Variance = E(X²) -

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃)

Four rotten oranges are mixed with 16 good ones. Three oranges are drawn one by one without replacement.

Let X denote the number of rotten oranges drawn

There are 4 rotten oranges present in 20 oranges

P(0) = = =

P(1) = = =

P(2) = = =

P(3) = = =

The probability distribution table is as follows,

Mean = E(X) = 0() + 1() + 2() + 3() = 0 + + + =

Mean = E(X) = =

= =

E(X²) = = P(x₁) + P(x₂) + P(x₃)

E(X²) = () + () + () + () = 0 + + + =

E(X²) = =

Variance = E(X²) - = – = =

Variance = E(X²) - =

Mean = E(X) =

Variance =

Given : Three balls are drawn simultaneously from a bag containing 5 white and 4 red balls.

To find : mean (𝓊) and variance (σ²) of X

Formula used :

Mean = E(X) =

Variance = E(X²) -

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃)

Three balls are drawn simultaneously from a bag containing 5 white and 4 red balls.

Let X be the number of red balls drawn.

P(0) = = =

P(1) = = =

P(2) = = =

P(3) = = =

The probability distribution table is as follows,

Mean = E(X) = 0() + 1() + 2() + 3() = 0 + + + =

Mean = E(X) = =

= =

E(X²) = = P(x₁) + P(x₂) + P(x₃)

E(X²) = () + () + () + () = 0 + + + =

E(X²) = =

Variance = E(X²) - = – = =

Variance = E(X²) - =

Mean = E(X) =

Variance =

Given : Two cards are drawn without replacement from a well-shuffled deck of 52 cards.

To find : mean (𝓊) and variance (σ²) of X

Formula used :

Mean = E(X) =

Variance = E(X²) -

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃)

Two cards are drawn without replacement from a well-shuffled deck of 52 cards.

Let X denote the number of face cards drawn

There are 12 face cards present in 52 cards

P(0) = = =

P(1) = = =

P(2) = = =

The probability distribution table is as follows,

Mean = E(X) = 0() + 1() + 2() = 0 + + = = =

Mean = E(X) =

= =

E(X²) = = P(x₁) + P(x₂) + P(x₃)

E(X²) = () + () + () = 0 + + =

E(X²) =

Variance = E(X²) - = – = =

Variance = E(X²) - =

Mean = E(X) =

Variance =

Given : Two cards are drawn with replacement from a well-shuffled deck of 52 cards.

To find : mean (𝓊) and variance (σ²) of X

Formula used :

Mean = E(X) =

Variance = E(X²) -

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃)

Two cards are drawn with replacement from a well-shuffled deck of 52 cards.

Let X denote the number of ace cards drawn

There are 4 face cards present in 52 cards

X can take the value of 0,1,2.

P(0) = =

P(1) = = =

P(2) = =

The probability distribution table is as follows,

Mean = E(X) = 0() + 1() + 2() = 0 + + = = =

Mean = E(X) =

= =

E(X²) = = P(x₁) + P(x₂) + P(x₃)

E(X²) = () + () + () = 0 + + =

E(X²) =

Variance = E(X²) - = – =

Variance = E(X²) - =

Mean = E(X) =

Variance =

Given : Three cards are drawn successively with replacement from a well – shuffled deck of 52 cards.

To find : mean (𝓊) and variance (σ²) of X

Formula used :

Mean = E(X) =

Variance = E(X²) -

Mean = E(X) = = x₁P(x₁) + x₂P(x₂) + x₃P(x₃)

Three cards are drawn successively with replacement from a well – shuffled deck of 52 cards.

Let X be the number of hearts drawn.

Number of hearts in 52 cards is 13

P(0) = =

P(1) = =

P(2) = =

P(3) = =

The probability distribution table is as follows,

Mean = E(X) = 0() + 1() + 2() + 3() = 0 + + + = =

Mean = E(X) =

= =

E(X²) = = P(x₁) + P(x₂) + P(x₃)

E(X²) = () + () + () + () = 0 + + + = =

E(X²) =

Variance = E(X²) - = – = =

Variance = E(X²) - =

Mean = E(X) =

Variance =

Given : Five defective bulbs are accidently mixed with 20 good ones.

To find : probability distribution from this lot

Formula used :

Five defective bulbs are accidently mixed with 20 good ones.

Total number of bulbs = 25

X denote the number of defective bulbs drawn

X can draw the value 0 , 1 , 2 , 3 , 4.

since the number of bulbs drawn is 4, n = 4

P(0) = P(getting a no defective bulb) = = =

P(1) = P(getting 1 defective bulb and 3 good ones) = =

P(1) = =

P(2) = P(getting 2 defective bulbs and 2 good one) =

P(2) = = =

P(3) = P(getting 3 defective bulbs and 1 good one) =

P(3) = =

P(4) = P(getting all defective bulbs) = = =

P(4) =

The probability distribution table is as follows,