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let's suppose we have 3 black balls and 2 red balls in a box. if we take out 4 balls WITH REPLACEMENT, which is the probability to get one red ball? my strategy would be to suppose that the first ball is red so 2/5 and then consider 3 black balls. (3/5*3/5*3/5). then multiply all 2/5* (3/5^3) -->8.6%
I know it is wrong, but why??
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let's suppose we have 3 black balls and 2 red balls in a box. if we take out 4 balls WITH REPLACEMENT, which is the probability to get one red ball? my strategy would be to suppose that the first ball is red so 2/5 and then consider 3 black balls. (3/5*3/5*3/5). then multiply all 2/5* (3/5^3) -->8.6%
I know it is wrong, but why??
Show more
Because the question asks about the probability of one ball being red, not the first ball being red. You can have the following 4 cases: RBBB BRBB BBRB BBBR
The probability of each case is 2/5*(3/5)^3, thus the overall probability is 2/5*(3/5)^3*4.
let's suppose we have 3 black balls and 2 red balls in a box. if we take out 4 balls WITH REPLACEMENT, which is the probability to get one red ball? my strategy would be to suppose that the first ball is red so 2/5 and then consider 3 black balls. (3/5*3/5*3/5). then multiply all 2/5* (3/5^3) -->8.6%
I know it is wrong, but why??
Show more
You have done it 80% correctly, but here is the mistake...
if you remember the basics of the probability of permutations or combinations, OR = + AND = *
Here the red ball can be chosen 1st time or 2nd time or 3rd time or 4th time that is the reason you will have to add 2/5 * (3/5)^3 4 times.... which is nothing but 4*2/5*(3/5)^3...
Hope this clarifies...
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.