For a large conference, attendees may carry a digital Badge and

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 · Accepted Answer

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.

From the information given, the total is 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 that (Badge yes, Access yes) = 96. Therefore, (Badge no, Access no) = 240 - 96 = 144.

Now we can calculate the required probabilities:

P(Access|Badge) = 96/96 = 1 (whenever a Badge is present, Access is also present)

P(neither) = (Badge no, Access no)/Total = 144/240 = 3/5

Answer:  P(Access|Badge) = 1, P(neither) = 3/5

I29-180

Aarya_San · Answer

I believe there's some information missing from this. We don't know anything about access pass holders except for the fact that they're also holding a badge.

P(Neither) = 240-96)/240 = 144/240 = 3/5

P(A|B) = No. of Access Pass holders/96

Now this value can be 1 or 2/3

reckx07 · Answer

Can some one tell me how P(A|B) can be incurred from the details that have been given in the question.

AbhishekP220108 · Answer

We are given, Total attendees = 240

Condition- An attendee has an Access Pass if and only if the attendee has a Badge. which means Badge ⇔ Access Pass. No one has one without the other.
 Exactly 96 attendees have at least one of these items. Since having a Badge implies having an Access Pass (and vice versa), “having at least one” really means “having both”.
So,

Number with Badge (and Access Pass) = 96
Number without either = 240 – 96 = 144

Now the probability:

P(Access Pass∣Badge)=P(Access Pass AND Badge)/P(Badge)

Since Badge ⇔ Access Pass, every Badge-holder has an Access Pass.

Thus, P(Access Pass AND Badge)=P(Badge), Therefore the probability = 1.

now the probability that a random attendee has neither is:
P(neither)=Number with neither/Total attendees=144/240=3/5

Dereno · Answer

Let’s consider a 4*4 matrix. Let us use the acronym DB for digital badge and AP for access passs.

1) An attendee has an Access Pass  if and only if  the attendee has a Badge.

If there is no Badge, then there exists no access pass, which means b = 0.

2) In a registry of 240 attendees = total attendees is 240.

3) exactly 96 attendees have at least one of these items.

atleast one = (DB only + AP only + DB and AP) = 96

Neither = 200 - 96 = 144.

d = 144.

We know that b = 0 and d = 144. Hence, b+d = 144.

Thus, (a+c) = 240 - (b+d) = 240 - 144 = 96

(a+c) = 96.

Since, all those with AP also has a DB.

C = 0

DB 
 NO DB 
 TOTAL 
 
  AP  
  a = 96  
  b = 0  
  a+b = a+0 = a = 96  
 
  NO AP  
  c = 0  
  d = 144  
  c+d = 0+144 = 144 
 
  TOTAL  
  a+c =  96   
  b +d =  144   
   240

P( Access / Badge ) = 96/96 =  1

P ( Neither ) = 144 / 240 = (48*3)/(48*5) =  3/5.

utkarshs2402 · Answer

I think P(Badge/Access) would be 1 and not vice versa. If the stem can be interpreted as "A person HAS Access if and only if they have Badge, then it makes sense P(Access/Badge) = 1".
Am I thinking correct?
I marked 1, because there was no other information to make any other choice.

Bunuel · Answer

The question is phrased correctly. “An attendee has an Access Pass if and only if the attendee has a Badge” means both always occur together. So P(Access|Badge) = 1 and P(Badge|Access) = 1 as well. Please check the OE above.

For a large conference, attendees may carry a digital Badge and

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	DB	NO DB	TOTAL
AP	a = 96	b = 0	a+b = a+0 = a = 96
NO AP	c = 0	d = 144	c+d = 0+144 = 144
TOTAL	a+c = 96	b +d = 144	240

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