Standard +0.8 This question requires students to recognize that if one side is x, the other is (10-x), then determine when max(x, 10-x) > 6, leading to a piecewise analysis. This goes beyond routine uniform distribution probability calculations and requires geometric insight about the rectangle constraint plus careful case analysis of which side is longer.
A rectangle has a perimeter of 20 cm. The length, \(x\) cm, of one side of this rectangle is uniformly distributed between 1 cm and 7 cm.
Find the probability that the length of the longer side of the rectangle is more than 6 cm long. [5]
total \(= \frac{1}{6} + \frac{1}{2} = \frac{2}{3}\)
\(1 - \frac{1}{3} = \frac{2}{3}\)
total \(= \frac{1}{6} + \frac{1}{2} = \frac{2}{3}\)
B1, M1, A1, M1dep B, A1
(5)
Methods 1 and 2: B1 for 6 and 4 (allow if seen on a diagram on x-axis); M1 for \(P(X > 6)\) or \(P(6 < X < 7)\); or \(P(X < 4)\) or \(P(1 < X < 4)\); or \(P(4 < X < 6)\); Allow \(\leq\) and \(\geq\) signs. A1: \(;or; \frac{1}{6}; \frac{1}{2}; \frac{2}{3}\) must match the probability statement. M1 for adding their "\(P(X > 6)\)" and their "\(P(X < 4)\)" or 1 - their "\(P(4 < X < 6)\)" dep on getting first B mark. A1 cao: \(\frac{2}{3}\) Method 3 Y~U[3, 9]: B1 for 6 with \(U[1,7]\) and 6 with \(U[3,9]\); M1 for \(P(X > 6)\) or \(P(6 < X < 7)\) or \(P(6 < Y < 9)\); A1: \(\frac{1}{6}; or; \frac{1}{2}\); must match the probability statement. M1 for adding their "\(P(X > 6)\)" and their "\(P(Y > 6)\)" dep on getting first B mark. A1 cao: \(\frac{2}{3}\)
| **Method 1** | **Method 2** | **Method 3** |
|---|---|---|
| $P(X > 6) = \frac{1}{6}$ | $P(4 < X < 6) = \frac{1}{3}$ | $P(X > 6) = \frac{1}{6}$ |
| $P(X < 4) = \frac{1}{2}$ | | $Y \sim U[3,9]$ $P(Y > 6) = \frac{1}{2}$ |
| total $= \frac{1}{6} + \frac{1}{2} = \frac{2}{3}$ | $1 - \frac{1}{3} = \frac{2}{3}$ | total $= \frac{1}{6} + \frac{1}{2} = \frac{2}{3}$ |
| B1, M1, A1, M1dep B, A1 | (5) | **Methods 1 and 2:** B1 for 6 and 4 (allow if seen on a diagram on x-axis); M1 for $P(X > 6)$ or $P(6 < X < 7)$; or $P(X < 4)$ or $P(1 < X < 4)$; or $P(4 < X < 6)$; Allow $\leq$ and $\geq$ signs. **A1:** $;or; \frac{1}{6}; \frac{1}{2}; \frac{2}{3}$ must match the probability statement. M1 for adding their "$P(X > 6)$" and their "$P(X < 4)$" or 1 - their "$P(4 < X < 6)$" dep on getting first B mark. **A1 cao:** $\frac{2}{3}$ **Method 3 Y~U[3, 9]:** B1 for 6 with $U[1,7]$ and 6 with $U[3,9]$; M1 for $P(X > 6)$ or $P(6 < X < 7)$ or $P(6 < Y < 9)$; **A1:** $\frac{1}{6}; or; \frac{1}{2}$; must match the probability statement. M1 for adding their "$P(X > 6)$" and their "$P(Y > 6)$" dep on getting first B mark. **A1 cao:** $\frac{2}{3}$ |
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A rectangle has a perimeter of 20 cm. The length, $x$ cm, of one side of this rectangle is uniformly distributed between 1 cm and 7 cm.
Find the probability that the length of the longer side of the rectangle is more than 6 cm long. [5]
\hfill \mbox{\textit{Edexcel S2 2010 Q3 [5]}}