From among a group of 2 men, 4 women and 6 children, 2 individuals are selected randomly. What is the probability that exactly 1 among the chosen are children?
prob(choosing child) = 6/12 = 1/2
prob(non-child) = 1- 1/2 = 1/2
prob(exactly one child in choosing two)
= C(2,1) (1/2)(1/2) = 1/2
or
If C stands for choosing child and N stands for non-child
we have these cases:
CC -- prob = (1/2)(1/2) = 1/4
CN -- 1/4
NC --1/4
NN -- 1/4
so prob is CN or NC = 1/4 + 1/4 = 1/2
To find the probability that exactly 1 among the chosen are children, we need to determine three probabilities: the probability of selecting 1 child and 1 non-child, the number of ways this can occur, and the total number of possible outcomes.
Let's start by calculating the probability of selecting 1 child and 1 non-child:
1. Calculate the probability of selecting 1 child:
There are 6 children in the group, so the probability of selecting 1 child is 6/12 (since there are a total of 12 individuals in the group).
2. Calculate the probability of selecting 1 non-child:
Since we need to select exactly 1 non-child, we can use the concept of complementary probabilities. There are 6 non-children in the group (2 men + 4 women), so the probability of selecting 1 non-child is 6/12.
3. Multiply the two probabilities calculated above to find the probability of selecting 1 child and 1 non-child:
P(1 child and 1 non-child) = (6/12) * (6/12) = 36/144 = 1/4
Next, we need to determine the number of ways this can occur:
Since there are 6 children and 6 non-children in the group, there are 6 possibilities for selecting a child and 6 possibilities for selecting a non-child. Multiplying these together gives us a total of 36 possible outcomes.
Finally, we need to determine the total number of possible outcomes:
The total number of individuals in the group is 2 men + 4 women + 6 children = 12. Therefore, the total number of possible outcomes is given by selecting any 2 individuals from a group of 12, which can be calculated using combinations:
Total possible outcomes = 12C2 = (12!)/(2!(12-2)!) = (12!)/(2!10!) = (12*11)/(2*1) = 66
Now, we can calculate the probability by dividing the number of favorable outcomes (1 child and 1 non-child) by the total number of outcomes:
P(exactly 1 child) = (Number of favorable outcomes) / (Total number of outcomes) = 36 / 66 = 6 / 11
Therefore, the probability that exactly 1 among the chosen are children is 6/11.