Abox contains 10 balls, of which 3 are red, 2 are yellow, and 5 are blue. Five balls are randomly selected with replacement. Calculate the probability that fewer than 2 of the selected balls are red?

Total number of balls = 10 
Red 3, Blue = 5 and Yellow =2 
3 Probability of getting red ball, p= 3/10
7 Probability of not getting red ball, q= 7/10 
Five balls are randomly selected, n = 5 
Now, probability that fewer than 2 of the selected balls 
are red,
$\begin{array}{l}P(X=0)+P(X=1)\\ {=}^5{C}_0{\left(\frac{3}{10}\right)}^0{\left(\frac{7}{10}\right)}^5{+}^5{C}_1{\left(\frac{3}{10}\right)}^1{\left(\frac{7}{10}\right)}^4\\ =\frac{7^5}{{10}^5}+\frac{3\times 7^4\times 5}{{10}^5}=\frac{7^4(7+15)}{{10}^5}=\frac{22\times 74}{{10}^5}\end{array}$
Probability = 0.5282.