Question

if two coins are tossed , find the expectation and the variance of the number of heads​

Answer 1

Answer:

S = {TT, TH, HT, HH}

Probability of getting head is : Let No. of heads is X

If X is 0 then P(X) is 1/4

If X is 1 then P(X) is 2/4

If X is 2 then P(X) is 1/4

Variance (X)  = E(X²) {E(X)}²

E(X) = ∑X × P(X)

       ⇒ 0 × 1/4 + 1 × 2/4 + 2 × 1/4

       ⇒ 1

E(X²) = ∑X² × P(X)

        ⇒ 0² × 1/4 + 1² × 2/4 + 2² × 1/4

        ⇒ 0 + 1/2 + 1

        ⇒ 1.5

∴ Variance (X) = 1.5 - 1²

                     = 0.5

Answer 2

Step 1: Given data

Two coins are tossed

Expectation of number of heads$=?$

Variance of number of heads$=?$

Step 2: Calculating expectation

Sample space, $S=$ { $TT, TH, HT, HH$}

Probability of getting a head$=\frac{1}{2}$

• if $X=0$, $P(X)=\frac{1}{4}$
• if $X=1$, $P(X)=\frac{2}{4}=\frac{1}{2}$
• if $X=2$, $P(X)=\frac{1}{4}$

Expectation, $E(X)=$ Σ$X\times P(X)$

$E(X)=0\times \frac{1}{4} +1\times \frac{2}{4} + 2\times \frac{1}{4}\\\\\\E(X)=1$

Step 3: Calculating variance

$E(X^{2} )=$Σ$X^{2} P(X)$

$E(X^{2} )=0^{2}\times \frac{1}{4} +1^{2} \times \frac{1}{2} + 2^{2} \times \frac{1}{4}\\\\\\E(X^{2} )=0+\frac{1}{2}+1 \\E(X^{2} )=1.5$

Variance is calculated using,

$E(X^{2} )-[E(X)]^{2} =1.5 -1^{2} =1.5-1=0.5$

Hence, expectation is $1$ and variance is $0.5$.

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