Question

Suppose a large jar contains eight red balls, six yellow balls, and six blue balls. Two balls are to be selected at random

from the jar, and the first ball selected will not be placed back into the jar.

(a) What is the probability that the first ball will be red and the second yellow?

(b) What is the probability that neither will be red?​

Answer 1

Answer:

(a) . PE=1/29 PE(of 2nd )=¹/²8

Step-by-step explanation:

(b). PE(of not red)=1-PE (of red)=1-6/28=5/28

Answer 2

Answer:

p(red balls) = 4/10

p(yellow balls) = 3/10

p(not red) = 3/5

Step-by-step explanation:

no. of red balls= 8

no. of yellow balls = 6

no. of blue balls = 6

total no. of outcomes = 8+6+6 = 20

(i) p(red ball) = 8/20

=> 4/10

p(yellow ball) = 6/20

=> 3/10

(ii) p(not red)= 20-(no. of red balls)

=> 20-8 = 12

=> 12/20 = 3/5