A spider eats 3 flies a day. Until the spider fill its quota, a fly has 50% chance of survival if it attempts to pass the web. Assuming 5 flies have already made the attempt to pass, what is the probability that the 6th fly will survive the attempt.
a) 0 %
b) 25 %
c) 50%
d) 75%
e) 100%
One way to solve it is to find out 1-P[FLY NOT EATEN] :
0 out of 5 fly eaten = [n!/k!(n-k!)]*1/2*1/2*1/2*1/2*1/2 = 1/32
1 out of 5 fly eaten = 5/32
2 out of 5 fly eaten = 10/32
Then find the probability of the 6th fly been eaten = 1/2*(1/32+5/32+10/32)= 1/4
So the prob of the 6th fly not been eaten is:
1-1/4=3/4=75%
=0.75 ——> 75%
The prob of any fly to pass thru is 1/2 regardless of how many have already been eaten.
C.
Ans is C
because here all the events are independent of each other. So their probability of survival will remain same.
c
I would go by Yo Dude and Hemant.
I think its 75%.
prob of spider being hungry is 1/2.
so total prob: P(spider hungry)xP(fly survives) + P(spider full)xP(fly survives)
(1/2 x 1/2) + (1/2 x 1) = 3/4 = 75%
bcoz when spider is full, fly will definitely survive.
This is what i had thought….
Take GMAT…Plz tell if its right concept or not??
This has got me thinking. I could think of the following two approaches to solve:
A) P(6th fly survives) = P(spider has eaten 3 flies from the first 5 flies) + P(fly survives)
= c(5,3)*(1/2)^5 + (1/2)
= (10/32) + (1/2)
= (5/16) + (1/2)
= 13/16 = 81.25 %
To me it seems that this is the right way to solve the problem. However none of the given answer options then match.
B) This is based on the assumption taht 3 flies are eaten from 6 flies.
Then:
P(6th fly eaten) = P(each fly being eaten)*P(fly eaten)
= (3/6)*(1/2) = 1/4
Therefore, P(fly survives) = 1 – P(6th fly eaten)
= 1-1/4 = 3/4 = 75%
I thin the former (A) method is right because there are no assumptions unlike the latter (B). However the latter gives the right answer (or the forced ‘right’ answer). I am confused.
I am not sure of Dan’s assumption that P (fly hungry) = 1/2
Ricky, could you please elaborate your reasoning.
I would appreciate to know where I am going wrong.
It’s C, the probability of of the 6th fly getting through is not dependent on anything else but the chance of survival.
I think the question is ambiguous. Does a fly survive even if its struck in the web and not eaten. In which case 75% is correct answer
@ hari
fly has 50% chance of survival only if spider has not eaten 3 flies (Its quota)…right??
here 2 cases arise:
1. Spider is hungry (has not eaten 3 flies yet)
2. Spider is full (has eaten 3 flies)
in first case, prob of fly survival is 50%,……….agree??
but in second case, fly will definitely survive. Therefore has prob. of 100%.
and also the probability that a spider is hungry or full is 50%.
because 2 cases are possible….hungry or full…..agree??
thus total prob is P(spider hungry)xP(fly survives) + P(spider full)xP(fly survives)
(1/2 x 1/2) + (1/2 x 1) = 3/4 = 75%
its 1/2 i.e. 50 %b as all events are independent of each other.
I think it’s more complicated than that…
The P of the first fly surviving is 1/2
The P of the second fly also surviving is 1/4
The P of all the first three flies surviving is 1/8
Before the third fly passes the P of the 6 fly surviving is 1/2…
Since there is a 1/8 chance of the first 3 flies all surving there is now a 1/8 P that the P of the 6th fly surviving is 100%
the P of all the first 4 flies surviving is 1/16
the P of all the 5 flies surviving eaten is 1/32
Since the P of the first flies surviving is 1/8 we add the P of all 4 flies and all 5 flies surviving 1/8 + 1/16+1/32 = 7/32 which give the 6th a P of 50% of surviving
25/32 is the P of at least 3 flies not surviving considering each one of those flies has a 1/2 P of surviving .
(7/32 X 1) + (25/32 X 1/2) = 60.94%
The writers of the answers made a mistake..
we dnt seem to have reached a conclusive answer to this even though the post is more than 6 months old..
Take GMAT team PLEASE HELP in identifying the right answer!
C
same here…take Gmat..can u please provide solution
this is ambiguous!
fo me it comes out to be 75%
I don’t think the question is clear enough…The question says “Until the spider fill its quota, a fly has 50% chance of survival if it attempts to pass the web”.
Does this mean that the fly will pass through the web with 100% chance of survival if the spider has eaten 3 flies?
Even so, the question says to assume 5 flies have passed but do we assume 5 flies have passed unhurt, or 5 flies have passed and 3 have been eaten because they will result in vastly different answers.
I completely agree with Yao. Here,the probability of survival of the 6th fly can be 0% if that is a fly among the 3 flies eaten up by spider.
P( survival) can also be 25%, 50% or 75%…….as others have explained the possibilities….
But,since it is mentioned in the question itself,that the probability of survival of each fly is 50%,I think the correct answer is
(c) 50%
Guys i got the answer from a different site…
but before i share that let me tell you an easy way in which i came to the conclusion..(the explanation given in the site is a bit complex)
Each fly has a 50% chance of being eaten…
so if the 6th fly also had a 50% chance of being eaten.
Since more than 3 flies flew into the web before this guy, the probability of him not getting eaten is more than 50% as the spider would have already had its fill. So only remaining ans choice is 75% and 100% and i go for 75%. Given below is the official explaination
Using the binomial distribution, the probability that n flies out of 5 have been eaten is combin(5,n)*(1/2)5, where combin(5,2)=5!/(n!*(5-n)!).
The probability that 0 flies were eaten is combin(5,0)*(1/2)5=1/32.
The probability that 1 fly was eaten is combin(5,1)*(1/2)5=5/32.
The probability that 2 flies were eaten is combin(5,2)*(1/2)5=10/32.
So the probability that the spider is still hungry is 1/32 + 5/32 + 10/32 = 16/32 = 1/2. The probability the spider is full is 1-1/2=1/2. Thus the probability of a successful attempt to pass is (1/2)*1 + (1/2)*0.5 = 0.75 .
Erricky taught it r8…thanx..!
C
50%
“survival if it attempts to pass the web” doesn’t matter if spidy has eaten or not. Spidy not going to go save it!
It is corporeal happiness to study you blog!