Question

Bob has a box of chocolates containing 6 chocolate covered almonds, 4 chocolate covered peanuts, and 8 plain chocolates. If the chocolates look identical, what is the probability that the first two chocolates Bob eats will have almonds?

Answer 1

The probability that the first two chocolates Bob eats will have almonds is 5/51 or 9.8% (nearest tenth).

Explanation:

Given that Bob has a box of chocolates containing 6 chocolate covered almonds, 4 chocolate covered peanuts, and 8 plain chocolates, the total number of chocolates in the box is:

`\implies \sf Total=6 + 4 + 8 = 18`

Probability Formula

`\boxed{\sf Probability\:of\:an\:event\:occurring = (Number\:of\:ways\:it\:can\:occur)/(Total\:number\:of\:possible\:outcomes)}`

Therefore, the probability that the first chocolate Bob eats will have almonds is:

`\implies \sf P(first\;almond)=(6)/(18)`

As Bob has eaten one almond chocolate there are now 5 chocolate covered almonds left and a total of 17 chocolates. Therefore, the probability that the next chocolate Bob eats will have almonds is:

`\implies \sf P(second\;almond)=(5)/(17)`

To find the total probability for two or more events, multiply the probabilities. Therefore, the probability that the first two chocolates Bob eats will have almonds is:

`\begin{aligned}\implies \sf P(first\;almond)\;and\;P(second\;almond) & =(6)/(18) * (5)/(17)\\\\ &=(30)/(306)\\\\&=(5)/(51)\\\\&=0.0980392...\\\\&=9.80392...\%\end{aligned}`

Therefore, the probability that the first two chocolates Bob eats will have almonds is 5/51 or 9.8% (nearest tenth).