Chocolate problem

**the prince** · April 10th 2010, 05:48 AM

there are 3 boxes with 10 chocolates in the first 5 of them are with strawberries , 10 chocolates in the second 4 of them are with strawberries 10 chocolates n the third and 3 of them are with strawberries
if we chose randomly one box and then randomly one chocolate how much is the probability to chose one that is with strawberries?

how to solve this? witch formula should i use ?
thanks for any reply

**e^(i*pi)** · April 10th 2010, 06:15 AM

Originally Posted by the prince

there are 3 boxes with 10 chocolates in the first 5 of them are with strawberries , 10 chocolates in the second 4 of them are with strawberries 10 chocolates n the third and 3 of them are with strawberries
if we chose randomly one box and then randomly one chocolate how much is the probability to chose one that is with strawberries?

how to solve this? witch formula should i use ?
thanks for any reply

It is better to find the chance of getting no strawberries and subtracting it from 1. If we let P(S) be the chance of getting a strawberry $P(S) = 1 - P'(S)$

To find P'(S) multiply the chance of not getting a strawberry in each box:

$P'(S) = \frac{5}{10} \times \frac{6}{10} \times \frac{7}{10}$

**Soroban** · April 10th 2010, 08:29 AM

Hello, the prince!

Very clumsy wording . . . Who wrote this problem?

There are 3 boxes with 10 chocolates in each of them.
Box A has 5 strawberries, box B has 4 strawberries, box C has 3 strawberries.

We chose randomly one box and then randomly choose one chocolate.
What is the probability that a strawberry is chosen?

There are three favorable outcomes.

. . $P(\text{Box A}\,\wedge\,\text{strawberry}) \;=\;\left(\frac{1}{3}\right)\left(\frac{5}{10}\ri ght) \;=\;\frac{5}{30}$

. . $P(\text{Box B}\,\wedge\,\text{strawberry}) \:=\:\left(\frac{1}{3}\right)\left(\frac{4}{10}\ri ght) \;=\;\frac{4}{30}$

. . $P(\text{Box C}\,\wedge\,\text{strawberry}) \;=\;\left(\frac{1}{3}\right)\left(\frac{3}{10}\ri ght) \;=\;\frac{3}{30}$

Therefore: . $P(\text{strawberry}) \;=\;\frac{5}{30} + \frac{4}{30} + \frac{3}{30} \;=\;\frac{12}{30} \;=\;\frac{2}{5}$

**mr fantastic** · April 11th 2010, 12:20 AM

Originally Posted by the prince

there are 3 boxes with 10 chocolates in the first 5 of them are with strawberries , 10 chocolates in the second 4 of them are with strawberries 10 chocolates n the third and 3 of them are with strawberries
if we chose randomly one box and then randomly one chocolate how much is the probability to chose one that is with strawberries?

how to solve this? witch formula should i use ?
thanks for any reply

Draw a tree diagram.

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