**Solve the following problem :**

A player tosses two coins. He wins ₹ 10 if 2 heads appear, ₹ 5 if 1 head appears, and ₹ 2 if no head appears. Find the expected value and variance of winning amount.

#### Solution 1

When a coin is tossed twice, the sample space is

S = {HH, HT, TH, HH}

Let X denote the amount he wins.

Then X takes values 10, 5, 2.

P(X = 10) = P(2 heads appear)= `1/ 4`

P(X = 5) = P(1 head appears) = `2/ 4= 1 /2`

P(X = 2) = P(no head appears) = `1/ 4`

We construct the following table to calculate the mean and the variance of X :

x_{i} |
P (x_{i}) |
xi P (x_{i}) |
x_{i}^{2} P (x_{i}) |

10 | `1/4` | `5/2` | 25 |

5 | `1/2` | `5/2` | 25/2 |

2 | `1/4` | `1/2`* | 1 |

Total |
1 |
5.5 |
38.5 |

From the table ∑x_{i} P(x_{i}) = 5.5, ∑x_{i}2 · P(x_{i}) = 38.5

E(X) = ∑x_{i}P(x_{i}) = 5.5

Var (X) = x_{i}^{ 2 }P(x_{i}) - [E(X)]^{2}

= 38.5 - (5.5)^{2}

= 38.5 - 30.25 = 8.25

∴ Hence, expected winning amount ₹ 5.5 and variance of winning amount ₹8.25

#### Solution 2

Let X denote the winning amount.

∴ Possible values of X are 2, 5, 10

Let P(getting head) = p = `(1)/(2)`

∴ q = 1 – p = `1 - (1)/(2) = (1)/(2)`

∴ P(X = 2) = P(no head) = qq= q^{2} = `(1)/(4)`

P(X = 5) = P(one head) = pq + qp = 2pq

= `2 xx (1)/(2) xx (1)/(2)`

= `(2)/(4)`

P(X = 10) = P(two heads) = pp = p^{2} = `(1)/(4)`

∴ The probability distribution of X is as follows:

X = x |
2 | 5 | 10 |

P(X = x) |
`(1)/(4)` | `(2)/(4)` | `(1)/(4)` |

Expected winning amount

= E(X) = \[\sum\limits_{i=1}^{3} x_i\text{P}(x_i)\]

= `2 xx (1)/(4) + 5 xx (2)/(4) + 10 xx (1)/(4)`

= `(2 + 10 + 10)/(4)`

= `(22)/(4)`

= ₹ 5.5

E(X^{2}) = \[\sum\limits_{i=1}^{3} x_i^2\text{P}(x_i)\]

= `(2)^2 xx (1)/(4) + (5)^2 xx (2)/(4) + (10)^2 xx (1)/(4)`

= `(4 + 50 + 100)/(4)`

= `(154)/(4)`

= 38.5

Variance of winning amount

= Var(X) = E(X^{2}) – [E(X)]^{2}

= 38.5 – (5.5)^{2}

= 38.5 – 30.25

= ₹ 8.25