In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
A. 21/46
B. 25/117
C. 1/50
D. 3/25
Could you please solve this problem with proper explanation?
Here at MHB, we normally don't provide fully worked solutions, but rather we help people to work the problem on their own. This benefits people much more, which is our goal.
Now, what you want to do here is to find the number of ways to choose 1 girl from 10 AND 2 boys from 15, then divide this by the number of ways to choose 3 children from 25. What do you find?
Now you just need to simplify, either by hand or with a calculator.
As probability is the ratio of the number favorable outcomes to the number of all outcomes, we are in this case dividing the number of ways to choose 1 girl from 10 AND 2 boys from 15 by the number of ways to choose 3 of the children from the total of 25. We are told that 3 children are selected at random, and we know there are 25 children by adding the number of boys to the number of girls. Thus, \(\displaystyle {25 \choose 3}\) is the total number of outcomes.
Thank you very much man! But I am still wondering why did we MULTIPLY # the number of ways to choose 1 girl from 10 # AND # 2 boys from 15 #, instead of ADDING these two operands?
We multiply because we are essentially applying the fundamental counting principle. When we require event 1 AND event 2 to happen, we multiply. When we require event 1 OR event 2 to happen, we add. Here we require both 2 boys AND 1 girl.
You see, for each way to obtain 2 boys, we have to account for all of the ways to obtain 1 girl. Or conversely, for each way to obtain 1 girl, we have to account for all of the ways to obtain 2 boys. The product of these two gives us all of the ways to get 2 boys and 1 girl.