Thread: Statistics Probability - DUE MONDAY

Relative Frequency of 0 centipede in a square = .45
Relative Frequency of 1 centipede in a square = .36
Relative Frequency of 2 centipede in a square = .14
Relative Frequency of 3 centipede in a square = .04
Relative Frequency of 4 centipede in a square = .01


1) Now suppose that 5 squares are chosen at random. Find the probability that 3 of the squares contain centipedes and 2 contain no centipedes. (hint: think binomial)

= 5!/(3! * 2!) * (.55^3) * (.45^2) = .34


2) If again, 5 squares are chosen at random, what is the probability that 3 or more squares contain centipedes?

= 5!/(3! * 2!) * (.55^3) * (.45^2) = .34

= 5!/(4! * 1!) * (.55^4) * (.45^1) = .21

= 5!/(5! * 0!) * (.55^5) * (.45^0) = .05


= .34 + .21 + .05 = .60







Is this correct? Please check my work. Thanks.

Hello funnyname7

Originally Posted by funnyname7

Relative Frequency of 0 centipede in a square = .45
Relative Frequency of 1 centipede in a square = .36
Relative Frequency of 2 centipede in a square = .14
Relative Frequency of 3 centipede in a square = .04
Relative Frequency of 4 centipede in a square = .01


1) Now suppose that 5 squares are chosen at random. Find the probability that 3 of the squares contain centipedes and 2 contain no centipedes. (hint: think binomial)

= 5!/(3! * 2!) * (.55^3) * (.45^2) = .34


2) If again, 5 squares are chosen at random, what is the probability that 3 or more squares contain centipedes?

= 5!/(3! * 2!) * (.55^3) * (.45^2) = .34

= 5!/(4! * 1!) * (.55^4) * (.45^1) = .21

= 5!/(5! * 0!) * (.55^5) * (.45^0) = .05


= .34 + .21 + .05 = .60







Is this correct? Please check my work. Thanks.

Yes, your work is good. Well done.

Just one thing, though: don't round off to 2 d.p. quite so soon. To 3 d.p. the answer is:

0.337 + 0.206 + 0.050 = 0.593

and you'll notice that this is 0.59 to 2 d.p.

Grandad

Thanks.