2 pair of dice. 1 red and 1 green rolled?

Question

Let E1 be the probability of the number on the freen due is 4 or less
Let E2 be the the probability of green die showing an odd number
Let E7 be the probability that the sum when two dies rolled is no greater than 5.

What is the probability: P(E1 n E2 n E7 | E1 u E2 u E7)

Greatly appreciate any help
Thank you!

1 PAIR of dice. NOT 2 1 red and 1 green.

Answer 1

1/5

P(E1 n E2 n E7 | E1 u E2 u E7) =
P(E1 n E2 n E7)/P(E1 u E2 u E7)

P(E1 n E2 n E7) = 1/6 since the green,red pairs that make this true are 1,1 1,2 1,3 1,4 3,1 3,2. Since there are 6 the probability is 6/36 = 1/6

P(E1 u E2 u E7) = 5/6 since this is true unless the green die is 6.

So the probability is 1/6 divided by 5/6 which is 1/5

Answer 2

Using just brute observation:

E1 = green 1, 2, 3 or 4
E2 = green 1, 3, 5

Since we want both E1 and E2 the only possibilities that satisfy both condition is:

green 1 or 3

E7 = sum of both dies less than or equal to 5, so we have the following possibilities:

green = 1 and red = 1, 2, 3, or 4
green = 3 and red = 1, or 2

Therefore we have 6 possibilities.

When we throw two die there are 6 x 6 = 36 total possibilities.

Therefore the probability of E1, E2 and E7 is 6/ 36 = 1/6

The union of E1 and E2 covers everything except green = 6.
E7 is fully encompassed in E1 union E2, so the probility is 5/6.

the answer to P(E1 n E2 n E7 | E1 u E2 u E7) = 1/6 / 5/6 = 1/5

Answer 3

employing merely brute remark: E1 = green a million, 2, 3 or 4 E2 = green a million, 3, 5 pondering the reality that we would like the two E1 and E2 the only prospects that fulfill the two subject is: green a million or 3 E7 = sum of each dies below or equivalent to 5, so we've the subsequent prospects: green = a million and purple = a million, 2, 3, or 4 green = 3 and purple = a million, or 2 subsequently we've 6 opportunities. whilst we throw 2 die there are 6 x 6 = 36 total prospects. as a effect the possibility of E1, E2 and E7 is 6/ 36 = a million/6 The union of E1 and E2 covers the comprehensive situation different than green = 6. E7 is easily encompassed in E1 union E2, so the probility is 5/6. the respond E1 u E2 u E7) = a million/6 / 5/6 = a million/5

Answer 4

The first part is pretty simple. Find the probability of each outcome and multiply them together.

E1 = 2/3
E2 = 1/2
E7 = 5/18

P(E1 n E2 n E7) = 2/3 * 1/2 * 5/18 = 5/54

It's been a long time since I've done any of this so I don't remember how to do the exclusions. Sorry. Hope this helps a little.

Answer 5

p(E1 u E2 u E7) = p(green<=4 OR green=odd OR sum<=5)
= p(green<=5 OR sum<=5) [combining E1 and E2]
= p(green<=5) [can't have sum<=5 without green being <=5]
= 5/6.

Now p(E1 n E2 n E7 n (E1 u E2 u E7))
= p(E1 n E2 n E7) = p(green<=4 AND green=odd AND sum<=5)
= p(green=1 or 3 AND sum<=5)
= 1/6 [The possibilities for (green,red) are (1,1), (1,2), (1,3), (1,4), (3,1), and (3,2) or 6 out of 36]

The conditional probability is (1/6) / (5/6) = 1/5.

Answer 6

E1 = 1/6
E2 = 3/6 = 1/2
E7 = 8/12 = 4/6 = 2/3
P(E1 n E2 n E7|E1 u E2 u E7)
this is as far as i go. good luck