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- October 22nd 2007, 06:32 PM #1

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## Last question

Candice own a chocolate shop. One of her most popular products is a box of 40 assorted chocolates, 5 of which contain nuts.

If a person selects two chocolates at random from the box, what is the probability that

i) both of the chocolate contains nuts

ii) at least one of the chocolates contains nuts

iii) only one of the chocolates contains nuts

iv) neither of the chocolates contains nuts?

Any help is MUCH MUCH MUCH appreciated

- October 22nd 2007, 06:55 PM #2

- October 22nd 2007, 07:14 PM #3

- October 22nd 2007, 08:30 PM #4
there are 40 chocolate candies to choose, 5 have nuts in them, so on the first selection we have a probability of 5/40 of selecting a box with nuts. now there are 39 candies left in the box, 4 of which have nuts, to select another box with nuts, the probability is now 4/39

what do we do with these two numbers? we want both events to happen, that is we want to select a candy with nuts AND select another with nuts.

[quote]

ii) at least one of the chocolates contains nuts

[quote]

at least one means one or more.

there are three ways for this to happen:

(1) you can select a candy with nuts on your first selection, and one without nuts on your second

(2) you can select one without nuts on your first selection and then one with nuts on your second

(3) you can select both candies that have nuts in them (this is what you did in the first part of the question).

since we have the desired outcome if (1) OR (2) OR (3) happens, we sum the probabilities.

what are the individual probabilities?

iii) only one of the chocolates contains nuts

iv) neither of the chocolates contains nuts?

(1) find the probability of NOT selecting a candy with nuts on you first selection AND then again NOT selecting a candy with nuts for your second selection. (use what i did in the first question as a clue of how to find each probability)

(2) you can calculate 1 - (probability of selecting two chocolates with nuts)