4-bad-apples-are-mixed-accidentally-with-20-good-apples-Obtain-the-probability-distribution-of-the-numbers-of-bad-apples-in-a-draw-of-2-apples-at-random-

Question Number 147906 by Odhiambojr last updated on 24/Jul/21

4 b a d a p p l e s a r e m i x e d a c c i d e n t a l l y

w i t h 20 g o o d a p p l e s . O b t a i n t h e

p r o b a b i l i t y d i s t r i b u t i o n o f t h e

n u m b e r s o f b a d a p p l e s i n a d r a w o f

2 a p p l e s a t r a n d o m

Answered by Olaf_Thorendsen last updated on 24/Jul/21

Let X be the probability of bad apples

in a draw . Then X is a random

variable with values 0, 1 and 2 .

Total apples are 4 Bad apples (B) +

20 Good apples (G) = 24 .

∙ P (X = 0) is the probability of no bad

apple in a draw of two apples .

P (X = 0) = \frac{C_{2}^{20}}{C_{2}^{24}} = \frac{20 \times 19}{24 \times 23} = \frac{95}{138}

∙ Similary P (X = 1) = \frac{C_{1}^{4} C_{1}^{20}}{C_{2}^{24}} = \frac{40}{138}

∙ P (X = 2) = \frac{C_{2}^{4}}{C_{2}^{24}} = \frac{4 \times 3}{24 \times 23} = \frac{3}{138}

Therefore, the probability distribution

of X is :

{\begin{cases} P (X = 0) = \frac{95}{138} \\ P (X = 1) = \frac{40}{138} \\ P (X = 2) = \frac{3}{138} \end{cases}