Question

A box 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 two of the selected balls are red.
A. 0.3601
B. 0.5000
C. 0.5282
D. 0.8369​

Answer 1

answer is A.

Hope it Help you .

Answer 2

Answer: C.0.5282

Step-by-step explanation:

This question can be solved by using Binomial Distribution.

As we know, Binomial distribution can only be applied to Dichotomy process.

Assume,

p - (probability of success) - Selecting Red ball

q - (probability of failure) - Selecting any other colour ball.

P[X<2] is what asked.

Therefore, we need to find for x = 0 and x = 1.

p = probability of selecting red ball = $\frac{3}{10}$

q = probability of selecting any other colour ball = $\frac{7}{10}$

Using the formula, nCr $p^{r}$ $q^{n-r}$ :

5C0× $(\frac{3}{10} )^{0}$ × $(\frac{7}{10}) ^{5}$ + 5C1× $(\frac{3}{10}) ^{1}$X $(\frac{7}{10}) ^{4}$.

Solving this, we get 0.5282.

Thank You. If any doubt, fell free to ask.