Question

1. Three six-sided dice are rolled. Let X be the sum of the dice. Determine the range of X and compute P(X = 18) and P(X ≤ 4).
2. An urn contains 5 red balls and 3 green balls. (a) Draw 3 balls with replacement. Let X be the number of red balls drawn. Determine the range of X and compute P(X = 3) and P(X 6= 1). (b) Draw 3 balls without replacement. Let Y be the number of red balls drawn. Determine the range of Y and compute P(Y = 3) and P(Y 6= 1)

Answer 1

Question 1

When three dices are rolled and obtained number has sumed.

P(x=18) is only possible when all three dices shows 6

(6,6,6)

So probability is 1/216

As 216 possible cases in 3 dice case.

For (X=< 4)

For P(x=0,1,2) is 0 because nothing possible

There are possible cases as (1,1,1) for P(x=3)

(1,2,1) (2,1,1) (1,1,2) for P(x=4)

So 4/216 is probability for P(=<4)

Question 2

3 balls are green

5 red balls

a) X = red balls

With replacement

P(X=3)= (5/8) (5/8) (5/8)

As there are 5 red balls out of 8 total balls and after having red we replace it into bag again and again having same probability 2 times more.

=125/512 = 0.244

P(X 6=1) i don't get this.... sorry

b)

Without replacement

Y= red balls

As 5 red balls out of 8 total balls after having red ball we keep it with themself and get another ball so bag has now 7 balls so probability of red ball is 4 out of 7 and then 3 out of 6 balls.

P(Y=3) = (5/8) (4/7) (3/6)

= 60/336

=0.1785

P(Y 6= 1)

I DON'T GET THESE 2 THINGS

SO SORRY ABOUT THIS

THANKS FOR POSTING

God bless you