A tool-hire depot stocks drills that are either cordless or corded. A drill is chosen at random from the depot. The probability that it is cordless is 0.58,...
Question 1 Report
A tool-hire depot stocks drills that are either cordless or corded. A drill is chosen at random from the depot. The probability that it is cordless is 0.58, and there are 87 cordless drills at the depot.
- Work out the total number of drills at the depot. (2)
- Work out the number of corded drills at the depot. (1)
Answer Details
This question tests reversing "probability = frequency / total" to find a total, then using the complement to find the remaining count.
(a) Since \( P(\text{cordless}) = \dfrac{\text{number cordless}}{\text{total}} \), the total number of drills at the depot is:
\[ \text{Total} = 87 \div 0.58 = 150 \] [2 marks]
(b) Every drill is either cordless or corded, so the number of corded drills is the total minus the number of cordless drills:
\[ \text{Corded drills} = 150 - 87 = 63 \] [1 mark]
This can be checked using the complement probability: \( (1 - 0.58) \times 150 = 0.42 \times 150 = 63 \).
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