Theory of Expectation :: Problems on Drawing Balls, Cards, Items, Products Probability Distribution

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An urn contains 7 white and 3 red balls. Two balls are drawn together, at random from this urn. Compute the probability that neither of them is white. Find also the probability of getting one white and one red ball. Hence compute the expected number of white balls drawn

Net Answers :
[Expectation: 1.4 ; Variance: 0.37 ; Standard Deviation: +0.608]

Solution

Total number of balls = 7 White + 3 Red

= 10

Total Number of possible choices (in drawing the two balls)
= Number of ways in which the two balls can be drawn from the total 12
= ¹⁰C_s

⇒ n =
10 × 9
2 × 1

= 45

Probabilty that

neither of them is white i.e. the two balls drawn would be "0 White and 2 Red"

⇒ P(0W2R) =
⁷C₀ × ³C₂
¹⁰C₂

White Red Total

Available 7 3 10

To Choose 0 2 2

Choices ⁷C₀ ³C₂ ¹⁰C₂

=

1 ×
3 × 2
2 × 1

45

=
3
45

=
1
15

One white and one red "1 White and 1 Red"

⇒ P(1W1R) =
⁷C₁ × ³C₁
¹⁰C₂

White Red Total

Available 7 3 10

To Choose 1 1 2

Choices ⁷C₁ ³C₁ ¹⁰C₂

=

7
1
×
3
1

45

=
21
45

=
7
15

Both white i.e. "2 White and 0 Red"

⇒ P(2W0R) =
⁷C₂ × ³C₀
¹⁰C₂

White Red Total

Available 7 3 10

To Choose 2 0 2

Choices ⁷C₂ ³C₀ ¹⁰C₂

=

7 × 6
2 × 1
× 1

45

=
21
45

=
7
15

"x" indicate the number of white balls obtained
[Since we are required to find the expected number of white balls, the variable would represent the number of white balls]
The number of white balls drawn would be

0 if "0 White and 2 Red" balls are drawn

1 if "1 White and 1 Red" ball are drawn

2 if "2 White and 0 Red" balls are drawn

⇒ The values carried by the variable ("x") would be either 0, 1, 2
⇒ "X" is a discrete random variable with range = {0, 1, 2}
"X" represents the random variable and P(X = x) represents the probability that the value within the range of the random variable is a specified value of "x"
Probability that the number of white balls drawn would be

0 ⇒ P(X = 0) = P(0W2R)

=
1
15

1 ⇒ P(X = 1) = P(1W1R)

=
7
15

2 ⇒ P(X = 2) = P(2W0R)

=
7
15

The probabilty distribution of "x" would be

x 0 1 2

P(X = x)
1
15

7
15

7
15

Calculations for Mean and Standard Deviations

x P (X = x) px
[x × P (X = x)] x² px²
[x² × P (X = x)]

0
1
15
0 0 0

1
7
15

7
15
1
7
15

2
7
15

14
15
4
28
15

Total 1
21
15

35
15

= 1.4 = 2.33

The expected number of white balls drawn

⇒ Expectation of "x"

⇒ E (x) = Σ px

= 1.4

Variance of the number of white balls drawn

⇒ var (x) = E (x²) − (E(x))²

⇒ var (x) = Σ px² − (Σ px)²

= 2.33 − (1.4)²

= 2.33 − 1.96

= 0.37

Standard Deviation of the number of white balls drawn

⇒ SD (x) = + √ Var (x)

= + √ 0.37

= + 0.608

Credit : Vijayalakshmi Desu

Theory of Expectation :: Drawing Balls, Cards, Items, Products