Paper 2 Section B Practice: Probability — Answers

Questions: Probability Drills.

Question 1

(a)

$$P(A\mid B)=\frac{0.30}{0.60}=\mathrm{0.5.}$$

[2]

(b)

$$P(A\cup B)=0.45+0.60-0.30=\mathrm{0.75.}$$

[2]

(c) If $A$ and $B$ were independent, $P(A\cap B)=P(A)P(B)=0.27$. Since $0.30\ne 0.27$, they are not independent. [2]

Question 2

(a)

$$P(\text{same colour})=\frac{5}{10}\frac{4}{9}+\frac{3}{10}\frac{2}{9}+\frac{2}{10}\frac{1}{9}=\frac{14}{45}.$$

[3]

(b)

$$P(\text{at least one red})=1-\frac{5}{10}\frac{4}{9}=\frac{7}{9}.$$

Also $P(\text{both red})=\frac{5}{10}\frac{4}{9}=\frac{2}{9}$. Therefore

$$P(\text{both red}\mid \text{at least one red})=\frac{2/9}{7/9}=\frac{2}{7}.$$

[3]

(c)

$$P(R_2\mid R_1)=\frac{4}{9},P(R_2)=\frac{1}{2}.$$

These are unequal, so $R_1$ and $R_2$ are not independent. [2]

Question 3

Let $D$ denote defective and $F$ denote flagged.

(a)

$$P(D)=0.65(0.018)+0.35(0.032)=\mathrm{0.0229.}$$

[3]

(b)

$$P(F)=P(F\mid D)P(D)+P(F\mid D^{\prime })P(D^{\prime })$$ $$=0.94(0.0229)+0.05(0.9771)=\mathrm{0.070381.}$$

Thus $P(F)\approx 0.0704$. [3]

(c) A flagged component from $Y$ may be defective or non-defective:

$$P(Y\cap F)=0.35[0.032(0.94)+0.968(0.05)]=\mathrm{0.027468.}$$

Hence

$$P(Y\mid F)=\frac{0.027468}{0.070381}\approx \mathrm{0.390.}$$

[2]