help with probability?

Question

Suppose you took a cenus of you school and found the followinf frequencies of blood types

blood type: A B AB O
Frequency: 78 22 8 92

if you chose 5 students at random, what would be the probability of each outcome?

a) finding 5 with type O

b) finding 5 with type B?

Answer 1

a) (92/200)(91/199)(90/198)(89/197)(88/196)
b) (22/200)(21/199)(20/198)(19/197)(18/196)

Answer 2

The chance that the first person chosen is type O is 92/200. The chance that th next person is O is 91/199 .
Therefore, the chance of 5 people from this group be all O is 92 x 91 x 90 x 89 x 88 / 200 x 199 x 198 x 197 x 196 = 0.193942+

Therefore, the chance of 5 people from this group be all B is 22 x 21 x 20 x 19 x 18 / 200 x 199 x 198 x 197 x 196 = 0.00001038550+

Answer 3

Okay, the question isn't that bad because it asks the probability of finding all with (something).

a) The probability that the first student chosen is type O is (92/200). Then, there are 91 type O's left out of 199 total students, so the probability of the second being a type O is (91/199). As you can see there is a pattern emerging:
the probability that third is type O (if first two were type O) is given by (90/198).
the probability that fourth is type O (if first three were type O) is given by (89/197)
the probability that fifth is type O (if first four were type O) is given by (88/196)

ANSWER:
(92/200)(91/199)(90/198)(89/197)(88/196) = 0.0193

b) Again, the method is similar to come up with the terms:
(22/200)(21/199)(20/198)(19/197)(18/196) = 0.000 0104

Answer 4

Hello

So there is a total of 200 students with blood types.

A.
1 student with O would be 92/200.
2nd student with O would be 91/199
3rd student with O would be 90/198
4th student with O would be 89/197
5th student with O would be 88/196.

Now multiply all the ratios to get .01939/1 or 5901255360/3.0467x10^11.

B. Now do the same procedure with Type B.
(22/200) * (21/199) * (20/198) * (19/197) * (18/196) = .000013855/1.

Hope this helps