<p>A teacher keeps a jar full of different flavored jelly beans on her desk and hands them out randomly to her class. But one greedy student likes only the licorice-flavored ones. One day after school, the student sneaks into the dark classroom and steals three jelly beans. If the jar has 50 beans in all15 licorice, 10 cherry, 20 watermelon, and 5 blueberrywhat is the probability that the student got at least one licorice-flavored bean?</p>
<p>?? can you explain how you came up with your answer please?</p>
<p>Let x C y be the choose function (binomial coefficient) where y objects are selected from x candidates.</p>
<p>In all, 50 C 3 combinations of beans can be selected. Of these, 15 * (49 C 2) combinations comprise licorice flavoured beans. So the answer is,</p>
<p>(15 * (49 C 2)) / (50 C 3)
= 0.9</p>
<p>I’m not entirely sure about this though.</p>
<p>I just subtracted the chance that the student picks no licorice at all (which is the opposite of picking at least one) from 1. </p>
<p>The chance that the bean you pick is not licorice is
(10 + 20 + 5)/50
= 7/10
= 0/7</p>
<p>0.7 x 0.7 x 0.7 (the chance that none of them are licorice) = 0.027.
Therefore, the chance that at least one is licorice is
1 - 0.027
= 0.973.</p>
<p>so? what is the correct answer? what did the answer key say?</p>
<p>I think you are wrong winslow… the probability of none being lico rice is not 0.7 because after u take the first bean the probability of the 2nd bean wouldn’t be “35/50” again… but “34/49” cuz of there is one bean less in the jar. and the 3rd would be “33/48”</p>
<p>So the probbility of it NOT being lico rce would be (35/50)<em>(34/49)</em>(33/48)= 0.334</p>
<p>The probability of it BEING lico rice then would be 1- 0.334= 0.666</p>
<p>But again I am no professional at probabilities so…</p>