Question

alberta diploma question
a child is allowed to have 2 treats from a bag containing 21 chocolates, 11 hard candies, and 10 gummy candies. all three types of treat are in packages of similar size and shape. since he cannot decide what he would like, the child reaches into the bag and randomly selects 2 treats, one at a time.
the probability that the child selects 1 chocolate and then 1 gummy candy, to the nearest thousandth, is
0.122
0.119
0.738
0.744
question help: message instructor

Explanation:

Step1: Calculate total number of treats

The total number of treats is $21 + 11+10=42$.

Step2: Calculate probability of selecting a chocolate first

The probability of selecting a chocolate on the first - draw is $P(\text{chocolate first})=\frac{21}{42}$.

Step3: Calculate probability of selecting a gummy candy second

After selecting a chocolate on the first draw, there are $41$ treats left. The probability of selecting a gummy candy on the second draw is $P(\text{gummy second})=\frac{10}{41}$.

Step4: Calculate combined probability

By the multiplication rule for independent events (in the sense of sequential non - replacement draws), the probability of selecting 1 chocolate and then 1 gummy candy is $P = \frac{21}{42}\times\frac{10}{41}=\frac{210}{1722}\approx0.122$.

Answer:

$0.122$