Collection of Maths Problems

Independence in rolling dice

Task number: 3541

• Variant

  \(n=6\), i.e. for a classic 6-sided die.

• Solution

  The probability space is \(\Omega=\{1,…,6\}\), \(P(\omega)=\frac16\) for all \(\omega\in\Omega\).

  The events are \(A=\{2{,}4,6\}\), \(B=\{4{,}5,6\}\) and \(A\cap B=\{4{,}6\}\) with probability \(P(A)=P(B)=\frac12\), \(P(A\cap B)=\frac13\ne \frac14=P(A)P(B)\).

• Answer

  The events \(A\) and \(B\) are dependent.

• Variant

  \(n=8\), i.e. an octahedron.

• Solution

  The probability space is \(\Omega=\{1,…,8\}\), \(P(\omega)=\frac18\) for all \(\omega\in\Omega\).

  The events are \(A=\{2{,}4,6{,}8\}\), \(B=\{5{,}6,7{,}8\}\) and \(A\cap B=\{6{,}8\}\) with probability \(P(A)=P(B)=\frac12\), \(P(A\cap B)=\frac14=P(A)P(B)\).

• Answer

  The events \(A\) and \(B\) are independent.

• Variant

  The parameter \(n\) is not specified exactly.

• Solution

  The probability space is \(\Omega=\{1,…,n\}\), \(P(\omega)=\frac1n\) for all \(\omega\in\Omega\).

  For \(n=4k\) the events are \(A=\{2{,}4,…,4k\}\), \(B=\{2k+1{,}2k+2,…,4k\}\) and \(A\cap B=\{2k+2{,}2k+4,…, 4k\}\) with probabilities \(P(A)=P(B)=\frac{2k}{4k}=\frac12\), \(P(A\cap B)= \frac{k}{4k}=\frac14=P(A)P(B)\).

  In this case the events \(A\) and \(B\) are independent.

  For \(n=4k+1\) the events are \(A=\{2{,}4,…,4k\}\), \(B=\{2k+1{,}2k+2,…,4k+1\}\) and \(A\cap B=\{2k+2{,}2k+4,…, 4k\}\) with probabilities \(P(A)=\frac{2k}{4k+1}\), \(P(B)=\frac{2k+1}{4k+1}\), \(P(A\cap B)= \frac{k}{4k+1} < \frac{k(4k+2)}{(4k+1)^2} =P(A)P(B)\).

  In this case the events \(A\) and \(B\) are dependent.

  For \(n=4k+2\) the events are \(A=\{2{,}4,…,4k+2\}\), \(B=\{2k+2,…,4k+2\}\) and \(A\cap B=\{2k+2{,}2k+4,…, 4k+2\}\) with probabilities \(P(A)=\frac{2k+1}{4k+2}=\frac12\), \(P(B)=\frac{2k+1}{4k+2}=\frac12\), \(P(A\cap B)=\frac{k+1}{4k+2}>\frac14=P(A)P(B)\).

  In this case the events \(A\) and \(B\) are dependent.

  For \(n=4k+3\) the events are \(A=\{2{,}4,…,4k+2\}\), \(B=\{2k+2,…,4k+3\}\) and \(A\cap B=\{2k+2{,}2k+4,…, 4k+2\}\) with probabilities \(P(A)=\frac{2k+1}{4k+3}\), \(P(B)=\frac{2k+2}{4k+3}\), \(P(A\cap B)=\frac{k+1}{4k+3} > \frac{(2k+1)(2k+2)}{(4k+3)^2} =P(A)P(B)\).

  In this case the events \(A\) and \(B\) are dependent.

• Answer

  The events \(A\) and \(B\) are independent only when \(4|n\).