Construct distribution then calculate probability

5 A couple plan to have at least one child of each sex, after which they will have no more children. However, if they have four children of one sex, they will have no more children. You should assume that each child is equally likely to be of either sex, and that the sexes of the children are independent. The random variable \(X\) represents the total number of girls the couple have.

1. Show that \(\mathrm { P } ( X = 1 ) = \frac { 11 } { 16 }\). The table shows the probability distribution of \(X\).

   \(r\)	0	1	2	3	4
   \(\mathrm { P } ( X = r )\)	\(\frac { 1 } { 16 }\)	\(\frac { 11 } { 16 }\)	\(\frac { 1 } { 8 }\)	\(\frac { 1 } { 16 }\)	\(\frac { 1 } { 16 }\)

2. Find \(\mathrm { E } ( X )\) and \(\operatorname { Var } ( X )\).

6 Three fair six-sided dice are thrown. The random variable \(X\) represents the highest of the three scores on the dice.

1. Show that \(\mathrm { P } ( X = 6 ) = \frac { 91 } { 216 }\). The table shows the probability distribution of \(X\).

   \(r\)	1	2	3	4	5	6
   \(\mathrm { P } ( X = r )\)	\(\frac { 1 } { 216 }\)	\(\frac { 7 } { 216 }\)	\(\frac { 19 } { 216 }\)	\(\frac { 37 } { 216 }\)	\(\frac { 61 } { 216 }\)	\(\frac { 91 } { 216 }\)

2. Find \(\mathrm { E } ( X )\) and \(\operatorname { Var } ( X )\).

6 Two bags contain coloured discs. At first, bag \(P\) contains 2 red discs and 2 green discs, and bag \(Q\) contains 3 red discs and 1 green disc. A disc is chosen at random from bag \(P\), its colour is noted and it is placed in bag \(Q\). A disc is then chosen at random from bag \(Q\), its colour is noted and it is placed in bag \(P\). A disc is then chosen at random from bag \(P\). The tree diagram shows the different combinations of three coloured discs chosen.
\includegraphics[max width=\textwidth, alt={}, center]{11316ea6-3999-4003-b77d-bee8b547c1da-05_858_980_573_585}

1. Write down the values of \(a , b , c , d , e\) and \(f\). The total number of red discs chosen, out of 3, is denoted by \(R\). The table shows the probability distribution of \(R\).

   \(r\)	0	1	2	3
   \(\mathrm { P } ( R = r )\)	\(\frac { 1 } { 10 }\)	\(k\)	\(\frac { 9 } { 20 }\)	\(\frac { 1 } { 5 }\)

2. Show how to obtain the value \(\mathrm { P } ( R = 2 ) = \frac { 9 } { 20 }\).
3. Find the value of \(k\).
4. Calculate the mean and variance of \(R\).