Question

URGENT PLEASE HELP MEEE Use the Venn diagram to calculate probabilities. Circles A, B, and C overlap. Circle A contains 12, circle B contains 11, and circle C contains 4. The overlap of A and B contains 5, the overlap of B and C contains 3, and the overlap of C and A contains 6. The overlap of the 3 circles contains 8. Which probabilities are correct? Select two options.

Answer 1

"Which probabilities are correct?" Without knowing which probabilities are given, this can't be answered precisely...

But at the very least, we can determine the probabilities of each of the events A, B, and C, as well as the probabilities of their intersections and unions (taken 2 at a time, or taking all 3 simultaneously).

The probability of any given event is equal to the proportion of the number of possible outcomes that make up that event to the total number of outcomes across all events.

Add up the numbers in each region of the Venn diagram to find the total:

12 + 11 + 4 + 5 + 3 + 6 + 8 = 49

Then the following probabilities come directly from the diagram:

$P(A)=\dfrac{12}{49}$

$P(B)=\dfrac{11}{49}$

$P(C)=\dfrac4{49}$

$P(A\cap B)=\dfrac5{49}$

$P(A\cap C)=\dfrac6{49}$

$P(B\cap C)=\dfrac3{49}$

$P(A\cap B\cap C)=\dfrac8{49}$

Using these probabilities and the inclusion/exclusion principle, we derive the probabilities of union: for any two events A and B,

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

which can be generalized for three events A, B, and C as

$P(A\cup B\cup C)=P(A)+P(B)+P(C)-(P(A\cap B)+P(A\cap C)+P(B\cap C))+P(A\cap B\cap C)$

So we find

$P(A\cup B)=\dfrac{12+11-5}{49}=\dfrac{18}{49}$

$P(A\cup C)=\dfrac{12+4-6}{49}=\dfrac{10}{49}$

$P(B\cup C)=\dfrac{11+4-3}{49}=\dfrac{10}{49}$

$P(A\cup B\cup C)=\dfrac{12+11+4-(5+6+3)+8}{49}=\dfrac{21}{49}=\dfrac37$

We can also get the probabilities of complements for free, since $P(A^C)=1-P(A)$, and conditional probabilities via $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$, etc.