Question

In a bag there are 10 balls. There are 4 red balls and the remaining balls are green balls removed at random and the color noted, the ball is replaced. Then a second ball is removed at random and the color is noted. P(1 red ball and 1 green ball picked)= 4/10*6/10=6/25 P(2 red balls picked given the first ball is red)

Name: in a bag there are 10 balls there are 4 red balls and the remaining balls are green balls removed at random and the color noted the ball is replaced then a second ball is removed at random a 16169
Uploaded: 2023-02-26T11:23:29-08:00
Duration: 1 min 11 s
Channel: Adi S
Description: in a bag there are 10 balls there are 4 red balls and the remaining balls are green balls removed at random and the color noted the ball is replaced then a second ball is removed at random a 16169

          In a bag there are 10 balls. There are 4 red balls and the remaining balls are green balls removed at random and the color noted, the ball is replaced. Then a second ball is removed at random and the color is noted.
P(1 red ball and 1 green ball picked)= 4/10*6/10=6/25
P(2 red balls picked given the first ball is red)

Added by Lisa C.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Adi S

02/26/2023

Step 1

Step 1: Calculate the probability of picking a red ball on the first draw P(picking a red ball on the first draw) = 4/10 = 2/5 Show more…

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In a bag there are 10 balls. There are 4 red balls and the remaining balls are green balls removed at random and the color noted, the ball is replaced. Then a second ball is removed at random and the color is noted. P(1 red ball and 1 green ball picked)= 4/10*6/10=6/25 P(2 red balls picked given the first ball is red)