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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. - Mathematics and Statistics

Sum

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.

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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 :

xi P (xi) xi P (xi) xi2 P (xi)
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 ∑xi P(xi) = 5.5, ∑xi2 · P(xi) = 38.5

E(X) = ∑xiP(xi) = 5.5

Var (X) = xiP(xi) - [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= q2 = `(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 = p2 = `(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(X2) = \[\sum\limits_{i=1}^{3} x_i^2\text{P}(x_i)\]

= `(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(X2) – [E(X)]2
= 38.5 – (5.5)2
= 38.5 – 30.25
= ₹ 8.25

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APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 8 Probability Distributions
Part I | Q 1.11 | Page 156
Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 7 Probability Distributions
Miscellaneous Exercise | Q 11 | Page 244
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