Need help!

westlondongaurav · #1 $Old$ November 6th, 2009, 02:18 PM

i was given this question to do in uni for my biology course, and i seem to be having problems solving it, i'm new to this site and was wondering if anyone could help out, cheers =)

1. A wood is divided into the 36 squares of equal size and the number of badger sets in each square was counted. 18 squares had no sets, 6 had 1 set, 6 had 2 sets, 3 had 3 sets, 2 had 4 sets and 1 had 7 sets. How many of the 36 sqaures might have been expected to have had 1 set in them if the sets were distributed at random in the wood? (10 marks)

awkward · #2 $Old$ November 6th, 2009, 06:52 PM

Quote:

Originally Posted by westlondongaurav $View Post$

i was given this question to do in uni for my biology course, and i seem to be having problems solving it, i'm new to this site and was wondering if anyone could help out, cheers =)

1. A wood is divided into the 36 squares of equal size and the number of badger sets in each square was counted. 18 squares had no sets, 6 had 1 set, 6 had 2 sets, 3 had 3 sets, 2 had 4 sets and 1 had 7 sets. How many of the 36 sqaures might have been expected to have had 1 set in them if the sets were distributed at random in the wood? (10 marks)

If I counted correctly, there are 42 badger sets in all.

Let
$X_i = 1$ if square i has exactly one badger set in it,
$= 0$ otherwise,
for i = 1, 2, 3, ... , 36.

Assuming the badger sets make their choices independently and that the squares are all equally likely to be chosen, the number of sets in a square follows a Binomial distribution with n = 42 and p = 1/36. So
$P(X_i = 1) = \binom{42}{1} \; (1/36)^1 (35/36)^{41}$ .
This is also $E(X_i)$ .

The expected value of the number of squares with exactly one badger set is then
$E(\sum_{i=1}^{36} X_i) = \sum_{i=1}^{36} E(X_i) = 36 \cdot \binom{42}{1} \; (1/36)^1 (35/36)^{41}$ , which is approximately 13.23.

We have used the theorem that E(X+Y) = E(X) + E(Y) above. It's important to realize that the theorem does not require that X and Y be independent. That is good for us, because the $X_i$ 's are not independent.

westlondongaurav · #3 $Old$ November 7th, 2009, 05:40 AM

thanks a lot for your help, i really appreciate it!! your explanation is great also, thanks

sorry i'm just having one little problem with that, isn't the equation for binomial distribution: P(X=r) = (nCr) pr (1-p)n-r ?
i don't see where 36 comes from when you multiply to get the answer on the final line.

okay sorry never mind i see now, we have 36 squares so we have to multiply the answer by 36, silly me