Example 4: HSC Mathematics paper 2001, Q7(b)

Onslo tries to connect to his internet service provider. The probability that he connects on any single attempt is 0.75.

(i)

What is the probability that he connects for the first time on his second attempt? 2 marks

(ii)

What is the probability that he is still not connected after his third
attempt? 1 mark

ANSWER

Denote $C_i$ as the event that connection takes place on the $i$th attempt.

Then probability of first connection at second attempt is

$$\begin{eqnarray*} P(\bar C_1 \cap C_2) & = & P(\bar C_1) \times P(C_2) \\ & = & 0.25 \times 0.75 \end{eqnarray*}$$

since the 2 events are independent.

The probability of failing at the first 3 attempts is

$$P(\bar C_1 \cap \bar C_2 \cap \bar C_3) = (0.25)^3$$

EXAMINERS' REMARKS, 2001

Question 7

(b)

Candidates who produced tree diagrams were much more successful in both parts. The most common m istake in (b)(i) was to assume that Onslo connected on his first attempt as well. Others had the correct probabilities for each attempt, but added them instead of multiplying. Candidates should note that it is important that they record their unsimplified answer in questions such as this, as an answer to (b)(ii) such as 0.016 without any supporting working does not necessarily convince the examiner that the candidate has correctly computed the probability.