Answer to Question #288272 in Statistics and Probability for ......

Question

1. A bag contains five yellow sweets, three green sweets and one purple sweet. A sweet is taken out

of the bag and replaced. Another sweet is taken out.

(a) Find the probability of neither sweet is green.

(b) Find the probability that the two sweets are different colours

Answer 1

(a) Total no. of sweets is 5+3+1=9

Probability of 1st sweet not being green = $\dfrac{6}{9}$

As the take sweet is replaced, the total no. of sweets will remain unchanged.

So Probability of 2st sweet not being green is also $\dfrac{6}{9}$ .

Therefore the probability of neither sweet is green $=\dfrac{6}{9}\times\dfrac{6}{9}=\dfrac{4}{9}$

(b) Probability that the two sweets are

(i) Yellow-Green and vice-versa $=2\times\dfrac{5}{9}\times\dfrac{3}{9}=\dfrac{10}{27}$

(ii) Purple-Green and vice-versa $=2\times\dfrac{1}{9}\times\dfrac{3}{9}=\dfrac{2}{27}$

(iii) Yellow-Purple and vice-versa $=2\times\dfrac{5}{9}\times\dfrac{1}{9}=\dfrac{10}{81}$

Hence net probability = $\dfrac{10}{27}+\dfrac{2}{27}+\dfrac{10}{81}=\dfrac{46}{81}$