There are 10 red and 20 blue balls in a box. A ball is chosen at random and it is noted whether it is red. The process repeats, returning the ball 10 times. Calculate the expected value and the standard deviation of this game.
Answers
Answer:
The probability of a red ball is 1/3 each draw with replacement. This is a binomial distribution with mean=np and variance=np(1-p)
E(x)=n/3
V(x)=n(1/3)(2/3)=2n/9
sd (x)=sqrt (2n)/3
Answer:
3.33; 1.49
Explanation:
The probability of a red ball is 1/3 each draw with replacement. This is a binomial distribution with mean=np and variance=np(1-p)
E(x) =n/3
V(x) =n(1/3) (2/3) =2n/9
sd(x) =sqrt (2n)/3
This isn't a binomial test for the reason that 1/3 function isn't met. The quantity of trials n = 2. There are most effective outcomes, a purple ball or a blue ball, of every trial. If we outline deciding on a purple ball as a success, then deciding on a blue ball is a failure. The possibility of having the primary ball purple is 510 in view that there are 5 purple balls out of 10 balls. So, p=510 and q=1−p=1−510=510. However, p and q do now no longer stay the identical for the second one trial. If the primary ball decided on is purple, then the possibility of having the second one ball purple is forty nine in view that there are most effective 4 purple balls out of 9 balls. But if the primary ball decided on is blue, then the possibility of having the second one ball purple is fifty nine in view that there are nevertheless 5 purple balls out of 9 balls.
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