A box contains four balls One red ball, one blue ball, and two green balls. We repeatedly draw a random ball from the box with r
eplacement until we have chosen at least oneball of each of three colors. Find E(Y), whereY is the number of draws.
1 answer:
Answer:
1/32
Step-by-step explanation:
First find the probability of picking red, blue and green balls respectively. The probability of picking any of the green balls are equivalent since only one ball of each colour should be picked. The detailed solution is found in the image attached. They are independent events so we multiply the probabilities.
You might be interested in
To solve this, the problem is solvable by elimination. So you see how the 5y and -5y cancel out? You basically add those two and you add the -7x and 4x and 10 and -10 leaving you with -3x=0. Dividing by each side gives you x=0. Then, plug 0 in for x in -7x+5y=10. Giving you
-7(0)+5y=10.
0+5y=10
5y=10
Divide by 5 on each side giving you y=2. So your answer is (0,2).
-7(0)+5y=10.
0+5y=10
5y=10
Divide by 5 on each side giving you y=2. So your answer is (0,2).