In a group of people, 32 are australian-born, 12 were born in asia and 7 were born in
europe. if 2 of the people are selected at random, find the probability that:
a they were both born in asia
b at least 1 of them will be australian-born
с both were born in europe.

a.      22/425

b.   368/425.

c.      7/425.

Step-by-step explanation:

There are a total of 32+12+7 = 51 people in the group.

a. Probability (First one is Asian) = 12/51 = 4/17.

Probability(second one is Asian) = 11/50.

So the probability that both are Asian is 4/17 * 11/50

= 22/425.

b.  Prob(First is Australian) = 32/51

    Prob( Second is not Australian) = 19/50.

Prob( first is Australian and Second is not) = 32/51 * 19/50.

Prob( First is not Australian and the second is) =  19/51 * 32/50.

Prob ( Both are Australian) = 32/51 * 31/50)

Prob( At least one is Australian = 32/51 * 19/50 +  19/51 * 32/50 + 32/51 * 31/50

= 368/425.

c.   Prob ( First choice born in Europe) = 7/51 and probability second was also born in Europe =  6/50

So the answer is 7/51 * 6/50 = 7/425.

2x^2+5x+17

your welcome. : )

step-by-step explanation :

we know that

the measurement of the outer angle is the semi-difference of the arcs which comprises

in this problem

m< xzy is the outer angle

substitute the values

the measure of angle xoy is equal to

> by central angle

substitute the value