Math Is Fun Forum

scar820

Member

Registered: 2007-03-09

Posts: 1

Can you help me with this problem?

The Sure Fire Pest Control Company maintains that their rattraps are the most effective in the market. The probability of catching a rat within one day is said to be 90%. If I set four traps, find the probability that;
a)      Three will catch rats
b)     At least two will catch rats
c)      No more than on will catch a rat
d)     What are the mean and standard deviation of this distribution?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Can you help me with this problem?

This is a binomial distribution. There are 4 trials, and each trials has a 0.9 probability of success.

a) Using the binomial distribution formula, P(3) = 4C3 * 0.9³ * 0.1 = 0.2916

b) Slightly more complicated this time, because of the 'at least'. This time, we need the probability that either 2, 3 or 4 rats will be caught. We can work these out separately and then add them together.

P(2) = 4C2 * 0.9² * 0.1² = 0.0486
P(4) = 4C4 * 0.9^4 = 0.6561

Therefore, P(≥2) = 0.0486+0.2916+0.6561 = 0.9963.

c) 'No more than one' is the opposite of 'at least 2'.
This means that we can just take the answer to b) away from 1.

P(≤1) = 1 - 0.9963 = 0.0037

d) In a binomial distribution with parameters (p,q), the mean is given by pq and the variance is given by pq(1-q).

Therefore, in this case, the mean is 4*0.9 = 3.6.
The variance is 4*0.9*0.1 = 0.36.
The variance is the square of the standard deviation, which means that the standard deviation is √(0.36) = 0.6.

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Why did the vector cross the road?
It wanted to be normal.