Thread: how to calculate this using binomial distribution

There are 5 balls in total, and 2 are black and 3 are red. If we choose with replacement, meaning we replace the ball we took out each time, what is the probability of choosing one ball of each color?

the binomial distribution says, for an experiment that can have either success of failure as an outcome



n is the number of trials, x is the number of successes, p is the probability of p occurring and q=1-p




How can I set that up to solve the question of what the chance of choosing one of each ball or choosing two of the same color?

Hey kingsolomonsgrave.

You have two choices: black or red. You need to formalize your question when it comes to choosing one ball: is it exactly one? At least 1? When you do this then just calculate the probability.

If I ask what is the probability of two red balls with replacement then i would get 3/5 times 3/5 if I choose two black balls it would be 2/5 times 2/5 and one of each color is 2/5 times 3/5 I think.

To get that using the formula I would say the probability of red is 3/5 so P=3/5 and so Q= 2/5 also x=2 so (n-x)=0 and we have two trials so n=2

so I get


2!/(2!0!) times (3/5)^2 times (2/5)^0

which is the same as 3/5*3/5 or 9/25

is that right?

for 1 red and 1 black I would get 2/5*3/5

but with the binomial formula we have

x= number of successes and red= success so x=1

P=3/5 and Q=2/5 and n=2 and(n-x)=1

2!/(1!*1!)(3/5)^1 (2/5)^1

which would be 2 times (3/5)(2/5) [twice as much as I thought it would be]

is this one right?

The binomial model is one where you have two choices (each independent of each other) with the same probability and the probability reflects getting x true successes and n - x failures (successes and failures can be whatever you want them to be: that's just a label).

So in line with what I said earlier: you need to figure out what events you are looking at (and you didn't answer my question before).

The binomial distribution with n trials models getting x successes (0 to n) and n - x failures (0 to n) in any order: if this is not the right model then pick another one that is right.

I assume this is the probability of one red and one black on two draws.

With replacement, the probability of red on the first draw is 3/5 and then the probability of black on the second draw is 2/5. The probability of black on the first draw is 2/5 and then the probability of red on the second is 3/5. The probability of "red then black" is (3/5)(2/5)= 6/25 and the probability of "black then red" is (2/5)(3/5)= 6/25. The probability of "red and black in either order" is 6/25+ 6/25= 2(6/25)= 12/25.

Without replacement,the probability of red on the first draw is 3/5 and then the probability of black on the second draw (because there are now only 4 balls) is 2/4. The probability of black on the first draw is 2/5 and then the probability of red on the second is 3/4. The probability of "red then black" is (3/5)(2/4)= 6/20= 3/10 and the probability of "black then red" is (2/5)(3/4)= 6/20= 3/10. The probability of "one red, one black in any order, without replacement" is 3/10+ 3/10= 2(3/10)= 6/10= 3/5.