Solution. Let \(X_i\) denote the outcome on the \(i^{\text {th}}\) dice

Let \(X\) denote the sum of outcome of the dice. Hence \[X = X_1 + X_2 + X_3.\] \begin {align*} E(X) & = E(X_1) + E(X_2) + E(X_3) = 3\left (\frac {7}{2}\right ) = \frac {21}{2} = 10.5. \end {align*} □

Solution. Define a random variable \(X_i\) as follows

\[ X_i = \begin {cases} 1, \, & \text {if the }\hspace {0.2cm}i^{\text {th}}\hspace {0.2cm}\text {man picks his own hat}\\ 0, & \text {otherwise}\\ \end {cases} \]

\( P(X_i = 1) = \dfrac {1}{N}\hspace {0.5cm}\) and \(\hspace {0.5cm} P(X_i = 0) = \dfrac {N - 1}{N}\hspace {0.5cm}\) then we have that \(E(X_i) = \dfrac {1}{N}.\)

Let  \(X = X_1 + X_2 + \cdots \cdots + X_n\) \begin {align*} E(X) & = E(X_1) + E(X_2) + \cdots \cdots + E(X_n) = N \times \dfrac {1}{N} = 1. \end {align*}

Find \(\, Var(X)\) \begin {align*} Var(X) & = Var(X_1 + X_2 + \cdots \cdots + X_n) = \sum _i Var(X_i) + 2\sum \sum _{i<j} Cov(X_i\, ,\, X_j). \end {align*}

\begin {align*} Var(X_i) & = \left (0 - \frac {1}{N}\right )^2 \times \frac {N - 1}{N} \, + \, \left (1 - \frac {1}{N}\right )^2\times \frac {1}{N}\\ & = \frac {1}{N^2}\left (\frac {N - 1}{N}\right ) \, + \, \left (1 - \frac {1}{N}\right )^2\cdot \frac {1}{N}\\ & = \frac {1}{N}\left [\frac {N - 1}{N^2} + 1 + \frac {1}{N^2} - \frac {2}{N}\right ]\\ & = \frac {1}{N}\left [-\frac {1}{N} + 1\right ]\\ & = \frac {1}{N}\left (\frac {N - 1}{N}\right )\\ & = \frac {N - 1}{N^2}. \end {align*}

For \(\, i < j\) \begin {align*} Cov(X_i\, , \, X_j) & = E\left [\left (X_i - \frac {1}{N}\right )\left (X_j - \frac {1}{N}\right )\right ]\\ & = E\left [X_iX_j - \frac {1}{N}\,X_i - \frac {1}{N}\, X_j + \frac {1}{N^2}\right ]\\ & = E\left (X_iX_j\right ) - \frac {1}{N}\cdot \frac {1}{N} - \frac {1}{N}\cdot \frac {1}{N} + \frac {1}{N^2}\\ & = E\left (X_iX_j\right ) - \frac {1}{N^2}\\ & = 1\cdot P(X_i = 1\, , \, X_j = 1) - \frac {1}{N^2}\\ & = P\left ( X_i = 1\big / X_j = 1\right )\, P\left (X_j = 1\right ) - \frac {1}{N^2}\\ & = \frac {1}{N - 1}\times \frac {1}{N} - \frac {1}{N^2}\\ & = \frac {1}{N(N - 1)} - \frac {1}{N^2}\\ & = \frac {N - (N - 1)}{N^2 (N - 1)}\\ & = \frac {N - N + 1}{N^2 (N - 1)}\\ & = \frac {1}{N^2 (N - 1)}. \end {align*}

\begin {align*} Var(X) & = \sum ^N_{i = 1} Var(X_i) + 2\sum ^{N-1}_{i<j}\sum ^N Cov(X_i\, ,\, X_j)\\ & = \sum ^N_{i=1}\frac {N - 1}{N^2} + 2\sum ^{N-1}_{i<j}\sum ^N\frac {1}{N^2(N- 1)}\\ & = 1. \end {align*} □

Solution. Define a random variable \(X_i\) such that \[X_i = \begin {cases} 0 & \text {if the}\hspace {0.2cm}i^{\text {th}}\hspace {0.2cm}\text {coupon is not there in the set of 10}\\ 1 & \text {if atleast one}\hspace {0.2cm}i^{\text {th}}\hspace {0.2cm} \text {coupon is there in the set of 10}\\ \end {cases} \]

\(P(X_i = 0) = \left (\dfrac {24}{25}\right )^{10}\) \begin {align*} P(X_i = 1) & = P(\text {atleast one}\hspace {0.2cm} i^{\text {th}}\hspace {0.2cm} \text {type is there in the set of 10})\\ & = 1 - P(X_i = 0)\\ & = 1 - \left (\dfrac {24}{25}\right )^{10}. \end {align*}

\[E(X_i) = 1 - \left (\frac {24}{25}\right )^{10}.\] Let \(\, X = X_1 + X_2 + \cdots \cdots + X_{25}\) \[ E(X) = 25\left ( 1 - \left (\frac {24}{25}\right )^{10}\right ).\] □