ii)

Method 1

There are 2 Ss and 3 Is so if exactly two of the three letters are the same they are either both Ss or both Is.

Both S's: P(S, S, Not S) + P(S, Not S, S) + P(Not S, S, S)

\begin{array}{rcl} P (exactly 2 Ss) & = & (\frac{2}{11} \times \frac{1}{10} \times \frac{9}{9}) + (\frac{2}{11} \times \frac{9}{10} \times \frac{1}{9}) + (\frac{9}{11} \times \frac{2}{10} \times \frac{1}{9}) \\ = & 3 (\frac{2 \times 1 \times 9}{11 \times 10 \times 9}) \end{array}

Probabilities multiplied [1]

Combinations Added [1]

Both I's: P(I, I, Not I) + P(I, Not I, I) + P(Not I, I, I)

\begin{array}{rcl} P (exactly 2 I s) & = & (\frac{3}{11} \times \frac{2}{10} \times \frac{8}{9}) + (\frac{3}{11} \times \frac{8}{10} \times \frac{2}{9}) + (\frac{8}{11} \times \frac{3}{10} \times \frac{2}{9}) \\ = & 3 (\frac{3 \times 2 \times 8}{11 \times 10 \times 9}) \end{array}

Probabilities multiplied [1]

Combinations Added [1]

Both S's or both I's: P(exactly 2 Ss) + P(exactly 2 Is)

\begin{array}{rcl} P (2 Ss or 2 I s) & = & 3 (\frac{2 \times 1 \times 9}{11 \times 10 \times 9}) + 3 (\frac{3 \times 2 \times 8}{11 \times 10 \times 9}) \\ = & 3 (\frac{18}{990}) + 3 (\frac{48}{990}) \\ = & \frac{54}{990} + \frac{144}{990} \\ = & \frac{198}{990} \end{array}

\frac{1}{5}

[1]
Unsimplified fraction or equivalent decimal accepted.

Method 2

There are 2 Ss and 3 Is so if exactly two of the three letters are the same they are either both Ss or both Is.

There are a number of different ways in which exactly 2 of the 3 letters could be the same, so it might help to consider a probability tree diagram.

In the following tree diagram, only pathways that lead to exactly 2 out of 3 letters being the same are shown.

P(exactly 2 the same) = P(S,S,Not S) + P(S, Not S, S) + P(I, I, Not I) + P(I, Not I, I) + P(Not S or I, S, S ) + P(Not S or I, I, I).

\begin{array}{rcl} P (exactly 2) & = & (\frac{2}{11} \times \frac{1}{10} \times \frac{9}{9}) + (\frac{2}{11} \times \frac{9}{10} \times \frac{2}{9}) + (\frac{3}{11} \times \frac{2}{10} \times \frac{8}{9}) \\ + (\frac{3}{11} \times \frac{8}{10} \times \frac{2}{9}) + (\frac{6}{11} \times \frac{2}{10} \times \frac{1}{9}) + (\frac{6}{11} \times \frac{3}{10} \times \frac{2}{9}) \\ = & \frac{18}{990} + \frac{36}{990} + \frac{48}{990} + \frac{48}{990} + \frac{12}{990} + \frac{36}{990} \\ = & \frac{198}{990} \end{array}

1 mark for any two combinations correct [2]
Probabilities multiplied [1]

Combinations Added [1]

\frac{1}{5}

[1]
Unsimplified fraction or equivalent decimal accepted.

Conditional Probability (CIE IGCSE Maths: Extended)

Exam Questions