5 In a game, a computer randemly chooses 12 shapes from 11 circles and 17 rectangles. The number of fectangles chosen) is denoted byR. (i) Show that \( \mathrm{P}(R=1)<\mathrm{P}(R=2) \). The number of rectangles available is now increased by \( r \). The computer randomly chooses 12 shapes from the 11 circles and \( (17+r) \) rectangles. The probability that 4 rectangles are chosen is now 15 times the probability that 3 rectangles are chosen. (ii) Find the value of \( r \).
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To show that \( \mathrm{P}(R=1) < \mathrm{P}(R=2) \): We can use the hypergeometric distribution. The probability of choosing \( R \) rectangles out of 12 shapes is given by the formula: \[ \mathrm{P}(R=k) = \frac{\binom{17}{k} \binom{11}{12-k}}{\binom{28}{12}} \] Where \( k \) is the number of rectangles chosen (either 1 or 2). Using this formula: For \( R = 1 \): \[ \mathrm{P}(R=1) = \frac{\binom{17}{1} \binom{11}{11}}{\binom{28}{12}} = \frac{17 \cdot 1}{\binom{28}{12}} = \frac{17}{\binom{28}{12}} \] For \( R = 2 \): \[ \mathrm{P}(R=2) = \frac{\binom{17}{2} \binom{11}{10}}{\binom{28}{12}} = \frac{\frac{17 \cdot 16}{2} \cdot 11}{\binom{28}{12}} = \frac{17 \cdot 16 \cdot 11}{2 \cdot \binom{28}{12}} \] Since \( 17 \cdot 16 \cdot 11 > 17 \), we see that: \[ \mathrm{P}(R=2) > \mathrm{P}(R=1) \] Now for the second part (ii), we set up the equation using the new total of rectangles: \[ \mathrm{P}(R=4) = \frac{\binom{17+r}{4} \binom{11}{8}}{\binom{28+r}{12}} \] \[ \mathrm{P}(R=3) = \frac{\binom{17+r}{3} \binom{11}{9}}{\binom{28+r}{12}} \] Given that \( \mathrm{P}(R=4) = 15 \cdot \mathrm{P}(R=3) \): This leads to: \[ \frac{\binom{17+r}{4} \cdot \binom{11}{8}}{\binom{28+r}{12}} = 15 \cdot \frac{\binom{17+r}{3} \cdot \binom{11}{9}}{\binom{28+r}{12}} \] Cancelling the common denominator \( \binom{28+r}{12} \) allows us to isolate the binomial coefficients: \[ \binom{17+r}{4} \cdot \binom{11}{8} = 15 \cdot \binom{17+r}{3} \cdot \binom{11}{9} \] This simplifies to: \[ \frac{\binom{17+r}{4}}{\binom{17+r}{3}} = 15 \cdot \frac{\binom{11}{9}}{\binom{11}{8}} \] Which leads us to recognize that \( \binom{a}{b} = \frac{a!}{b!(a-b)!} \). We calculate: \[ \frac{\binom{17+r}{4}}{\binom{17+r}{3}} = \frac{17+r-3}{4} \ok{= 15 \cdot 11} \] Thus, simplifying gives: \[ \frac{14+r}{4} = 165 \implies 14+r = 660 \implies r = 646 \] So the value of \( r \) is \( 646 \).
