Binomial distribution.

**Kane535** · May 15th 2010, 05:47 AM

Solve this problem:
Six calculators are bought. The probability of a calculator breaking down within two years is 0.25. Calculate the probability that within two years:

a) Exactly two of six calculators break
b) Three at most break
c) At least 3 break
d) Either two or three break

**Archie Meade** · May 15th 2010, 07:39 AM

Originally Posted by Kane535

Solve this problem:
Six calculators are bought. The probability of a calculator breaking down within two years is 0.25. Calculate the probability that within two years:

a) Exactly two of six calculators break
b) Three at most break
c) At least 3 break
d) Either two or three break

Hi Kane535,

this is a binomial situation in which a calculator breaks down within 2 years or doesn't.

$(p+q)^6=\binom{6}{0}p^6q^0+\binom{6}{1}p^5q^1+\bin om{6}{2}p^4q^2+\binom{6}{3}p^3q^3+\binom{6}{4}p^2q ^4+\binom{6}{5}pq^5+\binom{6}{6}p^0q^6$

where p=probability of a breakdown in 2 years, which is 0.25

q=probability the calculator is still functional after 2 years, which is 1-0.25=0.75.

$\binom{6}{n}p^{6-n}q^n$ is the probability of having exactly n functional calculators after 2 years.

(a)

$\binom{6}{4}p^2q^4=\binom{6}{2}p^2q^4$

(b)

At most 3 is 3 or less, which is 0, 1, 2 or 3 breakdowns

$\binom{6}{3}p^3q^3+\binom{6}{4}p^2q^4+\binom{6}{5} pq^5+\binom{6}{6}q^6$

(c)

At least 3 is 3, 4, 5 or 6 breakdowns

$\binom{6}{0}p^6+\binom{6}{1}p^5q+\binom{6}{2}p^4q^ 2+\binom{6}{3}p^3q^3$

(d)

Either 2 or 3

$\binom{6}{3}p^3q^3+\binom{6}{4}p^2q^4$

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