- You are given a
**ten piece**box of chocolate truffles. You know based on the label that six of the pieces have an orange cream filling and four of the pieces have a coconut filling. If you were to eat four pieces in a row, what is the probability that the**first two**pieces you eat have an orange cream filling and the**last two**have a coconut filling?

(given, O = orange, C = coconut) :

P(OOCC)

= P(1^{st} is orange) * P(2^{nd} is orange) * P(3^{rd} is coconut) * P (4^{th} is coconut)

= 6/10 * 5/9 * 4/8 * 3/7

= .6 * .556 * .5 * .429

= .0716

*Follow-up question:* If you were given an identical box of chocolates and again eat four pieces in a row, what is the probability that exactly **two** contain coconut filling?

- Step 1 involves a combinatorics problem of 4 choose 2 to determine how many combinations of oranges and coconuts we can obtain given 4 pulls from the box :

= _{4}C_{2}

= 4!/ (2! * (4-2)!)

= 24 / (2 * 2)

= 6

This is equivalent to the following six combinations:

CCOO, COCO, COOC, OCCO, OCOC, OOCC

- Step 2. In the first question above, we learned that the probability of pulling exactly 2 coconuts in 4 pulls from the box is the same at pulling exactly 2 oranges as well and we can see that the probability is the same for all 6 combinations of Os and Cs.

Therefore, the probability of pulling exactly 2 coconuts

= _{4}C_{2} [see step 1 above] * .0716

= 6 * .0716

= .4296