2015 H2 Mathematics Paper 2 Question 6

The Binomial Distribution

Answers

0.224{0.224}
Out of syllabus
0.824{0.824}

Full solutions

(i)

Let X{X} denote the r.v. of the number of red sweets in a small packet of 10{10}
XB(10,0.25)X \sim \textrm{B}\left(10, 0.25\right)
P(X4)=1P(X3)=0.224 (3 sf)  \begin{align*} \mathrm{P}(X \geq 4) &= 1 - \mathrm{P}(X\leq 3) \\ &= 0.224 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(ii)

Out of syllabus

(iii)

Let Y{Y} denote the r.v. of the number of red sweets in a large packet of 100{100}
YB(100,0.25)Y \sim \textrm{B}\left(100, 0.25\right)
P(Y30)=1P(Y29)=0.14954\begin{align*} \mathrm{P}(Y \geq 30) &= 1 - \mathrm{P}(Y\leq 29) \\ &= 0.14954 \end{align*}
Let W{W} denote the r.v. of the number of packets containing at least 30 red sweets out of 15{15}
WB(15,0.14954)W \sim \textrm{B}\left(15, 0.14954\right)
P(W3)=0.824 (3 sf)  \begin{align*} \mathrm{P}(W \leq 3) &= 0.824 \textrm{ (3 sf)} \; \blacksquare \end{align*}