Question

A box contains 2 strawberry yogurts, 4 vanilla yogurts and 6 cherry yogurts. Three yogurts are selected at random from the box. Calculate the probability that at least one of the selected yogurts is a cherry yogurt.

Answer 1

fraction
decimal or
percent?

Answer 2

1/2
.5
50%

Answer 3

6/12 =1/2. There's a 50% probability that a cherry yogurt will be selected.
Fraction: 1/2
Decimal: .50 or .5
Percent: 50%

Answer 4

is the probability
12 yogerts
6 are cherry
6/12
1/2
.5
50%

Answer 5

glad i could help!

Answer 6

is the selection of the 3 with or without replacement

Answer 7

you figure out if it is with/without replacement...then calculate

(probability that at least one of the selected yogurts is a cherry yogurt.)
=1-(probability that none of the selected yogurts is a cherry yogurt.)

Answer 8

Without replacement.

Answer 9

then the prob you select none of the cherry is

\[\frac{6}{12}\frac{5}{11}\frac{4}{10}=\frac{1}{11}\]

thus your probability is \[1=\frac{1}{11}=\frac{10}{11}\]

Answer 10

typo

\[1-\frac{1}{11}=\frac{10}{11}\]

Answer 11

That's not probability

Answer 12

?

Answer 13

Thanks anyway guys :)!

Answer 14

the answer is 10/11

Answer 15

this type of problem follows the hypergeometric distribution. you can write the above as

\[1-\frac{{6\choose 3}}{{12\choose3}}=\frac{10}{11}\]

Answer 16

you can also write it directly ..

\[\sum_{k=1}^{3}\frac{{6\choose k}{6\choose 3-k}}{{12 \choose 3}}=\frac{10}{11}\]

Answer 17

Thanks Zarkon, but thats the wrong way, Thanks anyway :)

Answer 18

it is not.