2017 H2 Mathematics Paper 2 Question 9

The Binomial Distribution

Answers

(i)

  • The probability of a light is faulty is the same for each light
  • Whether a light is faulty is independent of any other light

(ii)

0.632{0.632}

(iii)

0.000104{0.000104}

(iv)

0.458{0.458}

(v)

The event in part (iii) is a subset of the event in part (iv).
That is, if each box in a carton contains at least one faulty light, then there will be at least 20 faulty lights in the carton.
On the other hand, if there are at least 20 faulty lights in a carton, it does not necessarily mean that there are at least one faulty light in each box (eg, there may be two boxes with 10 faulty lights and 18 boxes with no faulty lights)

(vi)

69164{\frac{69}{164}}

(vii)

0.9408{0.9408}

(viii)

While the answer to (vii) suggests that the quick test gets the correct answer most (94.08% of the time), the answer to (vi) indicates that the quick test is not that worthwhile. In the event that the quick test identifies a light as faulty, there is still a significant probability of 69164{\frac{69}{164}} that the light is actually not faulty

Full solutions

(i)

  • The probability of a light is faulty is the same for each light   ■{\; \blacksquare}
  • Whether a light is faulty is independent of any other light   ■{\; \blacksquare}

(ii)

Let X{X} denote the r.v. of the number of faulty lights in a box of 12{12}
X∼B(12,0.08)X \sim \textrm{B}\left(12, 0.08\right)
P(X≥1)=1−P(X≤0)=0.63233=0.632 (3 sf)  ■\begin{align*} \mathrm{P}(X \geq 1) &= 1 - \mathrm{P}(X\leq 0) \\ &= 0.63233 \\ &= 0.632 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iii)

Let Y{Y} denote the r.v. of the number of red sweets in a large packet of 100{100}
Y∼B(20,0.63233)Y \sim \textrm{B}\left(20, 0.63233\right)
P(Y=20)=0.000104 (3 sf)  ■\begin{align*} \mathrm{P}(Y=20) &= 0.000104 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iv)

Let W{W} denote the r.v. of the number of faulty lights in a carton of 240{240}
W∼B(240,0.08)W \sim \textrm{B}\left(240, 0.08\right)
P(W≥20)=1−P(W≤19)=0.458 (3 sf)  ■\begin{align*} \mathrm{P}(W \geq 20) &= 1 - \mathrm{P}(W\leq 19) \\ &= 0.458 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(v)

The event in part (iii) is a subset of the event in part (iv).
That is, if each box in a carton contains at least one faulty light, then there will be at least 20 faulty lights in the carton.
On the other hand, if there are at least 20 faulty lights in a carton, it does not necessarily mean that there are at least one faulty light in each box (eg, there may be two boxes with 10 faulty lights and 18 boxes with no faulty lights)   ■{\; \blacksquare}

(vi)

P(not faulty∣identified as faulty)=P(not faulty∩identified as faulty)P(identified as faulty)=0.92×0.060.92×0.06+0.08×0.95=69164  ■\begin{align*} & \textrm{P}(\textrm{not faulty} \mid \textrm{identified as faulty}) \\ & = \frac{\textrm{P}(\textrm{not faulty} \cap \textrm{identified as faulty})}{\textrm{P}(\textrm{identified as faulty})} \\ &= \frac{0.92 \times 0.06}{0.92 \times 0.06 + 0.08 \times 0.95} \\ &= \frac{69}{164} \; \blacksquare \end{align*}

(vii)

P(correctly identified)=0.08×0.95+0.92×0.94=0.9408\begin{align*} & \textrm{P}(\textrm{correctly identified}) \\ & = 0.08\times0.95 + 0.92\times0.94 \\ &= 0.9408 \end{align*}

(viii)

While the answer to (vii) suggests that the quick test gets the correct answer most (94.08% of the time), the answer to (vi) indicates that the quick test is not that worthwhile. In the event that the quick test identifies a light as faulty, there is still a significant probability of 69164{\frac{69}{164}} that the light is actually not faulty   ■{\; \blacksquare}