Menu Close

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-th




Question Number 193292 by byaw last updated on 09/Jun/23
  (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?
$$ \\ $$(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?

Answered by MM42 last updated on 09/Jun/23
a)  if  A=rr ∪bb∪ww  the set is the same color then  thr sey?of non−uniformity in the face  A′=rb∪br∪rw∪wb∪bw∪wb  ⇒p(A)=(5/(10))×(3/9)+(4/(10))×(2/9)+(2/(10))×(4/9)=((29)/(90))  ⇒p(A′)=1−((29)/(90))=((61)/(90))  b)n→^6 g→^4 l→^3 g→^5 n⇒ans=6×4×3×5=360
$$\left.{a}\right)\:\:{if}\:\:{A}={rr}\:\cup{bb}\cup{ww}\:\:{the}\:{set}\:{is}\:{the}\:{same}\:{color}\:{then} \\ $$$${thr}\:{sey}?{of}\:{non}−{uniformity}\:{in}\:{the}\:{face} \\ $$$${A}'={rb}\cup{br}\cup{rw}\cup{wb}\cup{bw}\cup{wb} \\ $$$$\Rightarrow{p}\left({A}\right)=\frac{\mathrm{5}}{\mathrm{10}}×\frac{\mathrm{3}}{\mathrm{9}}+\frac{\mathrm{4}}{\mathrm{10}}×\frac{\mathrm{2}}{\mathrm{9}}+\frac{\mathrm{2}}{\mathrm{10}}×\frac{\mathrm{4}}{\mathrm{9}}=\frac{\mathrm{29}}{\mathrm{90}} \\ $$$$\Rightarrow{p}\left({A}'\right)=\mathrm{1}−\frac{\mathrm{29}}{\mathrm{90}}=\frac{\mathrm{61}}{\mathrm{90}} \\ $$$$\left.{b}\right){n}\overset{\mathrm{6}} {\rightarrow}{g}\overset{\mathrm{4}} {\rightarrow}{l}\overset{\mathrm{3}} {\rightarrow}{g}\overset{\mathrm{5}} {\rightarrow}{n}\Rightarrow{ans}=\mathrm{6}×\mathrm{4}×\mathrm{3}×\mathrm{5}=\mathrm{360}\: \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *