2019 H2 Mathematics Paper 2 Question 7

The Binomial Distribution

Answers

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

{0.102}

{0.991}

{45 p^2 (1-p)^{8}}

{0.0689}

The probability of a mug is faulty is the same for each mug

{\; \blacksquare}

Whether a mug is faulty is independent of any other mug

{\; \blacksquare}

F \sim \textrm{B}\left(50, 0.08\right)

\begin{align*} \mathrm{P}(F \geq 7) &= 1 - \mathrm{P}(F\leq 6) \\ &= 0.10187 \\ & = 0.102 \textrm{ (3 sf)} \; \blacksquare \end{align*}

Let

{D}

denote the r.v. of the number of days in which at least 7 faulty mugs were found out of of

{5}

days

D \sim \textrm{B}\left(5, 0.10187\right)

\begin{align*} \mathrm{P}(D \leq 2) &= 0.991 \; \blacksquare \end{align*}

Let

{S}

denote the r.v. of the number of faulty saucers out of a random sample

{10}

S \sim \textrm{B}\left(10, p\right)

\begin{align*} & \textrm{P}(S=2) \\ & = {10 \choose 2} p^2 (1-p)^{10-2} \\ & = 45 p^2 (1-p)^{8} \; \blacksquare \end{align*}

Let

{X}

and

{Y}

denote the r.v.s of the number of faulty mugs and faulty saucers respectively out of of

{2}

\begin{align*} X &\sim \textrm{B}(2, 0.08) \\ Y &\sim \textrm{B}(2, p) \\ \end{align*}

\begin{align*} & \textrm{P}(\textrm{at most } 1 \textrm{ faulty items}) \\ & = \textrm{P}(Y=0) \cdot \textrm{P}(X\leq 1) + \textrm{P}(Y=1) \cdot \textrm{P}(X = 0) \\ \end{align*}

\begin{gather*} 0.9936(1-p)^2 + 2p(1-p) \left( 0.8464 \right) = 0.97 \\ 0.9936(1-p)^2 + 1.6928 p(1-p) - 0.97 = 0 \; \blacksquare \end{gather*}

Since

{0 < p < 1,}

solving with a GC,

p=0.0689 \textrm{ (3 sf)} \; \blacksquare