I29-180

Question

For a large conference, attendees may carry a digital Badge and a venue Access Pass. An attendee has an Access Pass if and only if the attendee has a Badge. In a registry of 240 attendees, exactly 96 attendees have at least one of these items.
 
One attendee is selected at random. Select for  P(Access|Badge)  the probability that the attendee has an Access Pass given that the attendee has a Badge, and select for  P(neither)  the probability that the attendee has neither item. Make only two selections, one in each column.

Bunuel · Answer

Official Solution:

The possible states for an attendee are:
 
• (Badge yes, Access yes)
 
• (Badge no, Access no)
 
The states (Badge yes, Access no) and (Badge no, Access yes) are impossible, because “if and only if” means the two items always occur together. In particular, (Badge no, Access yes) is not possible because we are told that  an attendee has an Access Pass ... only if the attendee has a Badge . So, if someone does not have a Badge, he cannot have an Access Pass, otherwise it would violate the rule that every Access Pass must come with a Badge.
 
Total attendees = 240.
 
Since 96 attendees have at least one item, and under the “if and only if” condition this must mean they have both, we know (Badge yes, Access yes) = 96. Therefore, (Badge no, Access no) = 240 − 96 = 144.
 
Now we calculate the required probabilities:
 
P(Access|Badge) = 96/96 = 1
 
P(neither) = 144/240 = 3/5
 
 Answer:  P(Access|Badge) = 1, P(neither) = 3/5

Correct answer:

P(Access|Badge)  " 1 "

P(neither)  " \frac{3}{5} "

ritzysharma · Answer

Bunuel  I understood the logic for (Badge no, Access yes), but why is the case of (Badge yes, Access no) impossible? Some folks can have badge and not the access pass

Bunuel · Answer

The statement says:  An attendee has an Access Pass  if and only if  the attendee has a Badge.
 
Break it down:

The  “only if”  part means: if someone has an Access Pass, they must have a Badge. So (Badge no, Access yes) is impossible. 
 The  “if”  part means: if someone has a Badge, they must also have an Access Pass. So (Badge yes, Access no) is also impossible.

Together, “if and only if” forces both items to always appear together, either both present or both absent.

Hope it helps.

NetOrb · Answer

Thats a bad question, if and only if makes (badge no, access yes) impossible. 
badge yes, access no can still happen

someone can only have access if they have a badge is what the statement conveys 
If and only if you have badge you can have access.

but does having a badge automatically gives u access? no

Bunuel · Answer

The question is not bad, you are just missing the meaning of “if and only if.”

“An attendee has an Access Pass if and only if the attendee has a Badge” means both directions are true:

If someone has an Access Pass, then they have a Badge.
If someone has a Badge, then they have an Access Pass.

So “Badge yes, Access no” is also impossible. The two items must occur together.

By the way, this doubt has already been addressed above. Please review the discussion.

GameMoon · Answer

I like the solution - it’s helpful. Taught me the correct meaning of " If and only If". Quite insightful

FlatVase · Answer

why is it 96/96. As P(Access| Badge)=P(Both)/P(Badge); and P(Badge) meaning P(only badge)=0

Bunuel · Answer

P(Badge) means everyone who has a Badge, not “Badge only.”

Here, all 96 attendees with a Badge also have an Access Pass, so:

P(Access | Badge) = P(Both)/P(Badge) = 96/96 = 1.

P(Access\|Badge)	P(neither)
		\(0\)
		\(\frac{1}{5}\)
		\(\frac{2}{5}\)
		\(\frac{3}{5}\)
		\(\frac{2}{3}\)
		\(1\)

P(Access\|Badge)	P(neither)
		\(0\)
		\(\frac{1}{5}\)
		\(\frac{2}{5}\)
		\(\frac{3}{5}\)
		\(\frac{2}{3}\)
		\(1\)

I29-180

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