1. ## probability qustion

in a group 6 out of 10 men wear glasses and 5 out of 12 women wear glasses. if two people are chosen at random from the group what is the probability that they are either both men or both wear glasses.

probability that they are both men = 10/22*9/21 =15/77
probability that they both wear glasses. = 11/22 * 10/21 =5/21
or=+
15/77 + 5/21 =100/231 =not the right answer

2. ## Re: probability qustion

$P(A\cup B) = P(A)+P(B)-P(A\cap B)$
$P(\text{Both men or both wear classes}) = P(\text{Both men}) + P(\text{Both wear glasses}) - P(\text{Both men and both wear glasses})$
$=\dfrac{15}{77}+\dfrac{5}{21}-\dfrac{6}{22}\dfrac{5}{21} = \dfrac{85}{231}$

3. ## Re: probability qustion

Actually, I did not see that it is either/or. I just saw the or. That was my mistake. Since it cannot be both, then we cannot include any probabilities for both.

$P((A\cap B^c) \cup (A^c \cap B)) = P(A)+P(B)-2P(A\cap B) = \dfrac{70}{231}=\dfrac{10}{33}$

Another way of writing this:

$P(A\Delta B) = P(A\cup B) - P(A\cap B) = \left[P(A)+P(B)-P(A\cap B)\right] - P(A\cap B) = P(A)+P(B)-2P(A\cap B)$

4. ## Re: probability qustion

Originally Posted by SlipEternal
Actually, I did not see that it is either/or. I just saw the or. That was my mistake. Since it cannot be both, then we cannot include any probabilities for both.
$P((A\cap B^c) \cup (A^c \cap B)) = P(A)+P(B)-2P(A\cap B) = \dfrac{70}{231}=\dfrac{10}{33}$
Another way of writing this:
$P(A\Delta B) = P(A\cup B) - P(A\cap B) = \left[P(A)+P(B)-P(A\cap B)\right] - P(A\cap B) = P(A)+P(B)-2P(A\cap B)$
Not sure what A & B are above. Also realize the either/or meaning is not settled.
But I would read this as: A: (Two men) or B: (a man and a woman both wearing classes).
If read that way the two events are disjoint.

5. ## Re: probability qustion

Originally Posted by Plato
Not sure what A & B are above. Also realize the either/or meaning is not settled.
But I would read this as: A: (Two men) or B: (a man and a woman both wearing classes).
If read that way the two events are disjoint.
Or two women wearing glasses. Also, I described what A and B are in post #2.

6. ## Re: probability qustion

the answer at the back of the book is 85/231 so your first answer is the right one or is the book wrong?

7. ## Re: probability qustion

Originally Posted by markosheehan
the answer at the back of the book is 85/231 so your first answer is the right one or is the book wrong?
The problem is ambiguously worded, and it could be interpreted so that either my first answer or second answer could be correct. Plato seems to have read the problem according to my second interpretation, but also suggests that the "either/or" wording is ambiguous. If the book says that 85/231 is correct, clearly they were not interpreting either/or as implying disjoint events, but try to understand the distinction between the two. With that answer, you allow for the possibility that the chosen two are men who both wear glasses. But is that either/or? Either/or seems to imply that the two events are disjoint. They certainly are not. We can make them disjoint (Plato offered the an interpretation where they can be made into disjoint events). But, as written, they are not disjoint.

So again, I would say that the answer could be either of the answers I provided, depending on your interpretation, but the second answer (the one that does not match your book's answer) would be the one I would say is "more correct".

8. ## Re: probability qustion

Originally Posted by markosheehan
the answer at the back of the book is 85/231 so your first answer is the right one or is the book wrong?
$\dbinom{10}{2}+\dbinom{11}{2}-\dbinom{6}{2}=\dfrac{10\cdot 9}{2}+\dfrac{11\cdot 10}{2}-\dfrac{6\cdot 5}{2}=85$