(a) A man P has 5 red, 3 blue and 2 white buses. Another man Q has 3 red, 2 blue and 4 white buses. A bus owned by P is involved in an accident with a bus b...
Question 1 Report
(a) A man P has 5 red, 3 blue and 2 white buses. Another man Q has 3 red, 2 blue and 4 white buses. A bus owned by P is involved in an accident with a bus belonging to Q. Calculate the probability that the two buses are not of the same color.
(b) A man travels from Nigeria to Ghana by air and from Ghana to Liberia by ship. He returns by the same means. He has 6 airlines and 4 shipping lines to choose from. In how many ways can he make his journey without using the same airline or shipping line twice?
Answer Details
(a) P owns \(5R,3B,2W\) (total 10); Q owns \(3R,2B,4W\) (total 9). One bus of P meets one bus of Q. First find P(same colour):
\[P(\text{both R})=\tfrac{5}{10}\cdot\tfrac{3}{9}=\tfrac{15}{90},\quad P(\text{both B})=\tfrac{3}{10}\cdot\tfrac{2}{9}=\tfrac{6}{90},\quad P(\text{both W})=\tfrac{2}{10}\cdot\tfrac{4}{9}=\tfrac{8}{90}.\] \[P(\text{same})=\frac{15+6+8}{90}=\frac{29}{90}.\] \[P(\text{not same colour})=1-\frac{29}{90}=\frac{61}{90}.\](b) Outward journey: air Nigeria \(\to\) Ghana in 6 ways; ship Ghana \(\to\) Liberia in 4 ways. On the return he must not repeat the airline or shipping line, so: ship Liberia \(\to\) Ghana in \(4-1=3\) ways; air Ghana \(\to\) Nigeria in \(6-1=5\) ways.
\[\text{Total}=6\times 4\times 3\times 5=360\ \text{ways}.\]Download The App On Google Playstore
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