\(A\) and \(B\) are independent events. \(\text{P}(A) = x\), \(\text{P}(B) = x + 0.2\) and \(\text{P}(A \text{ and } B) = 0.15\). Find the value of \(x\).
Question 1 Report
\(A\) and \(B\) are independent events.
\(\text{P}(A) = x\), \(\text{P}(B) = x + 0.2\) and \(\text{P}(A \text{ and } B) = 0.15\).
Find the value of \(x\).
Answer Details
For independent events the probability that both occur is the product of the separate probabilities:
\[\text{P}(A\text{ and }B)=\text{P}(A)\times\text{P}(B)\]Substituting the given expressions turns a probability statement into a quadratic equation:
\[x(x+0.2)=0.15\][M1]
Expand and collect everything on one side:
\[x^{2}+0.2x-0.15=0\][M1]
Factorising, two numbers with product \(-0.15\) and sum \(0.2\) are \(-0.3\) and \(0.5\):
\[(x-0.3)(x+0.5)=0\][M1] oe (formula)
The two roots are \(x=0.3\) and \(x=-0.5\), but a probability cannot be negative, so
\[x=0.3\][A1] (\(x=-0.5\) rejected)
Rejecting the negative root with a brief reason is part of the answer, not an optional extra. Check the solution in context: \(\text{P}(A)=0.3\) and \(\text{P}(B)=0.5\), both valid probabilities, and \(0.3\times 0.5=0.15\) as required. The quadratic formula gives the same roots if the factorisation is not spotted, and multiplying the equation by 100 first, to give \(100x^{2}+20x-15=0\), avoids decimals if you prefer whole numbers.
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