Binomial complements and least n | CAIE 9709 Probability & Statistics 1 Notes

Probability & Statistics 1

Mathematics9709

Replace at least one by no successes

A complement is everything outside an event, so an event and its complement have total probability 1. If $X\sim B(n,p)$ and $q=1-p$ , then no successes means all $n$ trials fail.

P(X=0)=q^n

P(X\ge1)=1-q^n

Key idea

Use the complement when the direct event contains many possible counts but its opposite contains only one simple case.

The least n must fail once, then work

First value above 0.95

n = 12

0.931

fails

n = 13

0.945

fails

n = 14

0.956

works

n = 15

0.965

works

least n = 14

The least valid value must be the first bar above 0.95: n = 13 fails and n = 14 works.

Worked example

Find the least n for P(at least one success) > 0.95

Let the success probability on each trial be 0.2. Then $q=0.8$ .

1-0.8^n>0.95

0.8^n<0.05

Calculator powers are quickest. Check consecutive integers:

0.8^{13}=0.0550>0.05

0.8^{14}=0.0440<0.05

Answer

n=14

Key idea

Showing that 13 fails and 14 works proves that 14 is the first valid integer. Showing only that 14 works does not prove it is least.

Keep the inequality direction attached to the event

Common mistake

Do not stop at a calculator decimal for

n

. Round only after deciding whether the question needs the first integer above or below the boundary, then verify the adjacent integer.

Your turnIndependent check [4 marks]

Independent trials have success probability 0.15. Find the least number of trials needed for the probability of at least one success to be greater than 0.9.

Show worked answer

1-0.85^n>0.9

0.85^n<0.1

0.85^{14}=0.103>0.1

0.85^{15}=0.0874<0.1

Answer

n=15

Replace at least one by no successes

A complement is everything outside an event, so an event and its complement have total probability 1. If $X\sim B(n,p)$ and $q=1-p$ , then no successes means all $n$ trials fail.

P(X=0)=q^n

P(X\ge1)=1-q^n

Key idea

Use the complement when the direct event contains many possible counts but its opposite contains only one simple case.

The least n must fail once, then work

First value above 0.95

n = 12

0.931

fails

n = 13

0.945

fails

n = 14

0.956

works

n = 15

0.965

works

least n = 14

The least valid value must be the first bar above 0.95: n = 13 fails and n = 14 works.

Worked example

Find the least n for P(at least one success) > 0.95

Let the success probability on each trial be 0.2. Then $q=0.8$ .

1-0.8^n>0.95

0.8^n<0.05

Calculator powers are quickest. Check consecutive integers:

0.8^{13}=0.0550>0.05

0.8^{14}=0.0440<0.05

Answer

n=14

Key idea

Showing that 13 fails and 14 works proves that 14 is the first valid integer. Showing only that 14 works does not prove it is least.

Keep the inequality direction attached to the event

Common mistake

Do not stop at a calculator decimal for

n

. Round only after deciding whether the question needs the first integer above or below the boundary, then verify the adjacent integer.

Your turnIndependent check [4 marks]

Independent trials have success probability 0.15. Find the least number of trials needed for the probability of at least one success to be greater than 0.9.

Show worked answer

1-0.85^n>0.9

0.85^n<0.1

0.85^{14}=0.103>0.1

0.85^{15}=0.0874<0.1

Answer

n=15