Question

A bag contains 6 cherry candies, 3 orange candies, and 2 lemon candies. You reach in and take 3 pieces of candy at random. What is the probability that you have at least 1 lemon candy? Round to the nearest ten-thousandth when necessary.

Answer 1

at least one means not none

Answer 2

you can compute the probability you get no lemon candies, then subtract that from one
or you can compute the probability you get one lemon, and the probability you get two lemon, and add them up
which would you like to do?

Answer 3

the second option

Answer 4

ick

Answer 5

at first thought it was 2C1/11C3 but it didn't work out.

Answer 6

number of ways to choose 3 out of 11 is \(\binom{11}{3}\) which you may have seen written as \(_{11}C_3\) and is computed via
\[\frac{11\times 10\times 9}{3\times 2}=165\] and that will be your denominator

Answer 7

oh i see, ok since you have a good idea, lets do the easy way
the probability you get no lemon candies is
\[\frac{_9C_3}{_{11}C_3}\]

Answer 8

find that number, subtract it from 1, and you are done

Answer 9

these r the answer choices I have

Answer 10

0.1515
0.3362
0.4909
0.5758

Answer 11

ok I see

Answer 12

if you want the probability you have exactly one, you have to take in to account that it is one lemon, and two not lemon, so it would be
\[\frac{_9C_2\times 2}{_{11}C_3}\]

Answer 13

and then you can compute the probabilty you have exactly 2, which is
\[\frac{_9C_1}{_{11}C_3}\] or just \(\frac{9}{165}\) and then you have to add to the first number, but the other method is easier

Answer 14

thank you very much for ur help

Answer 15

Another approach, based on P(0) idea.

first pick, 2/11 lemon 1-2/11=0.82 probability of not lemon
second pick 2/10 lemon 1-2/10 = 0.80
third ick, 2/9 lemon 1-2/9 = 0.78

Multiply these probs get P(0) in three picks = 0.509
Thus prob at least one is 0.491

Answer 16

thank you