probability questions!!?

Question

I don't know how to do these, can someone explain please.

1. A bag contains 5 red, 6 blue, 2 white and 7 green balls. If 2 are seleccted at random (without replacement), find the probability of getting at least 1 red ball.

Answer: 17/38

2. In a group of people, 32 are Australian born, 12 were born in Asia, 7 were born in Europe. If 2 of the people are selected at random, find the probability that at least 1 of them will be Australian born.

Answer: 368/425

3. There are 34 men and 32 women at a party. Of these, 13 men and 19 woman are married. If 2 people are chosen at random, find the probability that both will be married.

Answer: 496/2145

what formula do you use?

Answer 1

When finding the probability of "at least one," it's efficient to find the complementary probability of "none," So in the first one, the probability of getting 0 red balls = 15/20 * 14/19 = 21/38. So P(at least 1 red) = 1-21/38 = 17/38

Same idea with the second question, the probability of "at least one Australian born" = 1 - P(0 Australians). P(0 Australians) = 19/51*18/50 = 57/425, so the answer here is 1-57/425 = 368/425

#3) There are 32 married people out of a total of 66 people at the party. The men and women is not really relevant because the question is just about picking 2 people at random. So it's 32/66*31/65 = 496/2145

Answer 2

1. Total number of balls is 20
You can draw a red ball the first time, in which case you don't care what the 2nd draw is. That probability is 5/20. There are 15/20 cases where you don't draw a red ball on the first draw. With 19 balls left, the probability of drawing a red ball is 5/19. Combining these,
P(1 red) = (5/20) + (15/20)(5/19)
P(1 red) = (5*19 + 15*5)/(20*19)
P(1 red) = (95 + 75)/380
P(1 red) = 170/380 = 17/38

2. This one works the same as the 1st one. The total number of people is 51. Let P(A) represent the probability of at least one Australian born:
P(A) = 32/51 + (19/51)(32/50)
P(A) = (32*50 + 19*32)/(51*50)
P(A) = 32(50 + 19)/2550
P(A) = 16(69)/1275
P(A) = 16(23)/425
P(A) = 368/425

3. There are 66 people. 32 are married
P(2 married) = (32/66)(31/65)
P(2 married) = (16/33)(31/65)
P(2 married) = 496/2145

Answer 3

1. there are 5 + 6 + 2 + 7 = 20 balls
there are 6 + 2 + 7 = 15 balls that are NOT RED

how may ways to get 2 balls? 20 * 19 = 380
how many ways to get 2 balls that are not red = 15 * 14 = 210

so the probability of getting no red balls from d drawings without replacement of 20 balls is 210/380 = 21/38

the probability of getting at least one red ball is (38 - 21)/38
= 17 / 38

2.same concept.
32 + 12 + 7 = 51 people
12 + 7 = 19 Not Australian

51*50 = 2550 ways of getting two people from a group of 51
19*18 = 342 ways of getting two people from 19 non australians

probability of getting no australians = 342/2550
probability of getting at least one australians =
(2250 - 342)/2550 =
2208/2250 =
368 / 425

3.
there are 34 + 32 = 66 people
and there are 13 + 19 = 32 married people

there are 66*65 = 4290 ways of getting two peoplein a group of 66
there are 32*31 = 992 ways of getting two married people in a group of 32 married poeple

so there are 992 / 4290 = 496 / 2145 ways of getting two married people in a group of 66 as decsribed

Answer 4

D