s*******y
发帖数: 105 | 1 A spider must eat three flies a day to survive. (It doesn't need to eat more
than three.) The spider has a 50% chance of catching any fly that comes
near its web. What are the spider's chances of survival, given that five
flies have already come near its web today?
I calculated in the following way:
0.5^5 + 5 * 0.5^1*0.5^4+10*0.5^2*0.5^3=0.5
The answer is 0.75
what 's wrong in my calculation? Thanks |
p*****k
发帖数: 318 | 2 supercury, 0.5 is correct for the problem as you stated.
i think the "correct" answer 0.75 you mentioned is to another
commonly seen (GMAT) question, e.g., see problem 169 on
http://mathproblems.info/group9.html |
o**o
发帖数: 3964 | 3 simply because a binomial distribution is symmetric |
s********t
发帖数: 31 | 4 Guess better to add 2 points to make the problem clear:
1. Exactly 5 fly visit during the whole day.
2. Even if the spider is full it will continue to catch fly with 50% success
rate. |
A*****8
发帖数: 73 | 5 Can you tell me the definition of chance of survival?
Is the possibility that spider eat no less than 3 flies?
If eat 3 fliers can survive, then no need to calculate, from the symmetry of binomial tree, the possibility no less than 3 is equal the possibility less than 3, two P total is 1, so the result is 1/2. |
