Question

For questions 15-20, consider a box with 3 red and 5 blue balls. 15. If one ball is drawn at random, what is the probability that the color is red? The probability that you pick a red ball is 3/8. 3 red, 5 blue 3+5=8 Therefore, 3/8. https://canvas.park.edu/files/5516027/download?download_frd=1&verifier=6xdC2s8YTdrrfONH1TZPJUK9jRKzh2UmtaqIbg9y 16. If two balls are drawn at random, what is the probability that both are red? Utilizing the tree... P (R X R)=(3/8) (2/7) 17. Two balls are drawn at random. If it is known that the first one is red, then what is the probability that the second one is red? 18. If two balls are drawn at random, what is the probability that they have the same color? 19. If two balls are drawn at random, what is the probability that they have different colors? 20. If three balls are drawn at random, what is the probability that at least one of them is blue?

Answer 1

Answer)

We have 3 red balls and 5 blue balls

Probability of selecting a red ball = 3/8

Probability of selecting a blue ball = 5/8

So,your 15, 16 both are correct

17)

Probability is given by favorable/total

Total was 8

But now one is already selected so, we are now left with 7

And we know that first one is red

And we had 3 red

But now we are left with 2 red

So, required probability is 2/7

18)

There are two possibilities

First both are red and second both are blue

Both are red = (3/8)*(2/7)

And both are blue = (5/8)*(4/7)

= 26/56

= 13/28

19)

We know that sum of all the probabilities is = 1

So, p(same colour) - p(different colours) = 1

So, p(differenr colour) = 1 - (13/28) = 15/28

20)

Answer)

Answer)

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 5/8

N = number of trials = 3

R = desired success = at least 1 = p(1)+p(2)+p(3)

We know that sum of all the probabilities = 1

So, p(atleast 1) = 1 - p(0)

= 0.947265625