A family has 4 children: Ariel, Bob, Chris, and Diana. Assuming that all birth orders are equally likely.
Find the probability that Ariel is older than Diana.
Find the probability that Ariel is older than Diana, if it is given that Ariel is older than Bob.
Find the probability that Ariel is older than Diana, if it is given that Chris is older than Diana.
Find the probability that Ariel is older than Diana, if it is given that Diana is older than Bob.

A) 0.5

B) 0.375

C) 0.375

D) 0.167

Step-by-step explanation:

Let's call

a: Ariel

b: Bob

c: Chris

d: Diana

There are 4*3*2*1 = 24 total combinations to order the 4 children

A) Ariel is older than Diana in the following 12 cases:

a b c d

a b d c

a c b d

a c d b

a d b c

a d c b

b a c d

b a d c

b c a d

c a b d

c a d b

c b a d

Then, the probability is 12/24 = 0.5

B) The condition "Ariel is older than Diana, if it is given that Ariel is older than Bob" is satisfied in the six first cases and last three cases of before.

Then, the probability is 9/24 = 0.375

C) Ariel is older than Diana, if it is given that Chris is older than Diana, in the following cases:

a b c d

a c b d

a c d b

a d c b

b a c d

b c a d

c a b d

c a d b

c b a d

Then, the probability is 9/24 = 0.375

D) Ariel is older than Diana, if it is given that Diana is older than Bob, in the following cases:

a c d b

a d b c

a d c b

c a d b

Then, the probability is 4/24 = 0.167

use this formula:

p(a\b) = p(a and b)/p(b).

take it from here.

i think the answer is seven, for if i'm being useless °ω°

step-by-step explanation: