A bag 4

A bag contains 18 balls that differ only in colour: 11 are blue and 7 are red. Two balls are picked one after the other without replacement. Find the probability that both are:
(i) blue,
(ii) the same colour,
(iii) different colours.

Final Answer:

a =  0.3595
b =  0.4967
c =  0.5033

Step-by-step explanation:

$a=\frac{11}{18}\cdot \frac{10}{17}=\frac{11\cdot 10}{18\cdot 17}=\frac{110}{306}=\frac{55}{153}=0.3595$

$b=\frac{11}{18}\cdot \frac{10}{17}+\frac{7}{18}\cdot \frac{6}{17}=\frac{76}{153}=0.4967$

$c=1-b=1-\frac{76}{153}=\frac{1\cdot 153}{153}-\frac{76}{153}=\frac{153}{153}-\frac{76}{153}=\frac{153-76}{153}=\frac{77}{153}=0.5033$

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