To solve the question, we will be using a Venn Diagram. Let's list down what's given first:
855 drank alcohol
657 smoked cigarettes
192 used illegal drugs
418 both drank alcohol and smoked cigarettes
109 drank alcohol and used illegal drugs
116 smoked cigarettes and used illegal drugs
88 engaged in all three
and 251 engaged in none of these
We will draw a Venn Diagram with 3 circles, one for each vice. Then we will start with the intersection of all three and the area outside of the circles (for those who do not engage in vices).
Because we know how many engaged in all 3 vices, we can deduct that number from those who belong to the intersection of 2 vices since they have already been counted in that too.
418 - 88 = 330 drank and smoked, but not used drugs
109 - 88 = 21 drank and used drugs, but not smoked
116 - 88 = 28 smoked and used drugs but not drank
Then, we subtract the numbers in each circle from the total for each vice.
855 - (330 + 88 + 21) = 416 drank only
657 - (330 + 88 + 28) = 211 smoked only
192 - (88 + 21 + 28) = 55 used drugs only
To find the total number of surveyed students, we just need to add all the numbers in the Venn Diagram.
416 + 330 + 88 + 21 + 211 + 28 + 55 + 251 = 1,400
Therefore, there were 1,400 students who were surveyed.
Answer: n = 10.
Step-by-step explanation:
In the bag, we have n counters.
4 of the counters are red.
the rest are blue, then we have:
(n - 4) blue counters.
Now, the probability that Ross takes a blue counter from the bag is equal to the quotient between the number of blue counters (n - 4) and the total number of counters, n
Then the probability is:
p1 = (n - 4)/n
Now he draws another, and it must be blue again, then we can calculate the probability in the same way as above, but he already take a blue counter, so the number of blue counters is (n - 5) and the total number of counters is (n - 1)
The probability of this event is:
p2 = (n - 5)/(n - 1)
The joint probability (the probability that Ross takes two blue counters) is equal to the product of the individual probabilities, and we know that this is equal to 1/3, then we have the equation:
1/3 = ( (n - 4)/n)*((n - 5)/(n - 1))
Now let's solve this for n.
n*(n - 1)/3 = (n - 4)*(n - 5)
(n^2 - n)/3 = n^2 - 4*n - 5*n + 20
n^2 - n = 3*(n^2 - 9*n + 20)
n^2 - n = 3*n^2 - 27*n + 60
0 = (3*n^2 - n^2) - 27*n + n + 60
0 = 2*n^2 - 26*n + 60
The two solutions of this equation can be found with Bhaskara's equation:
Then the two solutions are:
n = (26 - 14)/4 = 3
This is not an option, because we know for sure that we have 4 red counters, then this option can be discarded.
The other solution is:
n = (26 + 14)/4 = 40/4 = 10
Then we have n = 10, 10 counters in total.