Question

There are three boxes each containing 3 pink and 5 yellow balls and also there are 2 boxes each containing 4 pink and 2 yellow balls. A yellow ball is selected at random. Find the probability that yellow ball is from a box of the first group?

Answer 1

Answer:

let the pink balls be p1,p2,p3 and p4

and the yellow balls be y1,y2,y3,y4 and y5

Step-by-step explanation:

since a yellow ball is t be picked randomly

Therfore,S ={p1,p2,p3,y1,y2,y3,y4,

therefore,n(S)=

let A be the event that there are yellow balls to be selected

Therefore n(A)=5

by defination

$p(a ) = \frac{n(a)}{n(s)}$

therfore,p(a)=5/8

Answer 2

⇒  Here, we can see there are total 5 boxes.

⇒  P(E1) be the probability selecting  three boxes.

∴  P(E1)=3/5

⇒ P(E2) be the probability selecting other two boxes.

∴  P(E2)=2/5

⇒  Probability of selecting yellow ball from three boxes P(A/E1)=5/8

⇒  Probability of selecting yellow ball from two boxes P(A/E2)=2/6

 Now, the probability that yellow ball is from a box of first group 

P(E1)×P(A/E1)/P(E2)×P(A/E2)+P(E1)×P(A/E1)

(3/5*5/8)/(3/5*5/8)+(2/5*2/6)=45/16

ans: 45/16