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## Statistics and probability

### Course: Statistics and probability > Unit 7

Lesson 7: Multiplication rule for independent events- Sample spaces for compound events
- Sample spaces for compound events
- Compound probability of independent events
- Probability of a compound event
- "At least one" probability with coin flipping
- Free-throw probability
- Three-pointer vs free-throw probability
- Probability without equally likely events
- Independent events example: test taking
- Die rolling probability with independent events
- Independent probability
- Probabilities of compound events
- Probabilities involving "at least one" success
- Probability of "at least one" success

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# Probabilities involving "at least one" success

AP.STATS:

VAR‑4 (EU)

, VAR‑4.A (LO)

, VAR‑4.A.4 (EK)

, VAR‑4.E (LO)

, VAR‑4.E.2 (EK)

## Example 1 : Defective chips

A manufacturer of processing chips knows that 2, percent of its chips are defective in some way.

Suppose an inspector randomly selects 4 chips for an inspection.

**Assuming the chips are independent, what is the probability that at least one of the selected chips is defective?**

Lets break this problem up into smaller pieces to understand the strategy behind solving it.

## Example 2: Surgical implants

Surgeries involving implants sometimes result in the patient's body rejecting the implant. A certain surgery has a rejection rate of 11, percent. The rest of the patients successfully accept the implant.

Assume that the results for each patient are independent.

## Example 3: Free-throws

Esther is a basketball player who makes 75, percent of the free-throws she attempts. Assume that the results of each shot are independent.

## Generalizing the strategy

In general, we can use these strategies:

or similarly,

## Want to join the conversation?

- So, P(at least 1 success)=1−P(all failures)

then what will be P(at least 2 success) ?(16 votes)- Assume there are n independent trials, each with constant success probability p.

P(at least 2 successes) = P(at least 1 success) - P(exactly 1 success)

= 1 - P(all failures) - P(exactly 1 success)

= 1 - P(all failures) - (n choose 1)P(1 success followed by n-1 failures)

= 1 - P(all failures) - np(1 - p)^(n - 1), or equivalently 1 - (1 - p)^n - np(1 - p)^(n - 1).

Have a blessed, wonderful day!(39 votes)

- Hi. Where can I find a video or explanation for why in Example 1. Problem C you subtracted the probability from one to get at least one not defective?

I would guess it's the remainder probability because they have to add up to 1. But then if shouldn't you subtract 0.98 to the 3rd from 1?(4 votes)- 0.98 to the 3rd would tell you the probability of getting 3 good chips out of 3 picks.

Whilst we'd like to figure out what is the % of getting**at least**1 bad chip out of 4 picks, hence you'd have to sum up probabilities of all following cases:

(B - bad, G - good)

B B B B

----------

G B B B

B G B B

B B G B

B B B G

----------

G G B B

G B G B

G B B G

B G G B

B G B G

B B G G

----------

G G G B

G G B G

G B G G

B G G G

All of the above are "at least one bad chip" cases. Total sum of their probabilities is approx. 0.07763183999999998(26 votes)

- Here to find at least 1 defective chip, why P(SSSD) = 0.98 * 0.98 * 0.98 * 0.02 is not used. Is there any problem with this method?(2 votes)
- P(SSSD) is the probability that just the last chip selected is defective, and no others are defective. On the other hand, the probability that at least 1 chip is defective is the probability that 1, 2, 3, or all 4 of the chips are defective, which may or may not mean that the last chip selected is defective.

So P(SSSD) alone fails to take into account several other possibilities. Even 4*P(SSSD) still fails to take into account the possibility that 2, 3, or all 4 chips are defective.

Because of these many possibilities, it is much easier to do 1 - P(SSSS) = 1 - (0.98)^4 to find P(at least 1 defective).(17 votes)

- Why is it

1 - P(all successes) = P(at least one failure)

and not

P(failure)*P(rest are successful)

eg. I have a 2/3 chance of finding a maple tree in the woods by my house if I just walk randomly from tree to tree, so the chase that I would stumble upon at least one maple as I go about 10 trees would be:`P(Maple) * P(not a maple)^9 = P(at least one maple)`

(2/3) * (1/3)^9 = 0.00003

but if I do what Sal says I get:

`1 - P(all not maples) = P(at least one maple)`

1 - (1/3)^10 = 0.99998

And that makes more sense because I'm almost definitely not going to go from tree to tree and NOT encounter a maple, but*why*does this work?

Edit (-ish):

I had a theory, that you would find the probability of each event where you found at least one maple and you added them, and I got (to my even greater confusion):`sum(((1/3)^10-x)*((2/3)^x), x, 1, 10) = 0.03467`

Is there a reason that summation isn't getting the right answer either-- or am I doing it wrong?

Thanks in advance!(3 votes)- You missed, that not only the first tree but also any other tree could be a maple tree.

In my opinion, your calculation returns the probability of finding exactly one maple-tree at the beginning, disregarding, that it could also be part of another sequence

e.g. (0 = not found, 1 = found)

1000000000

0100000000

...

0000000001

That's imo why your estimate is too low

Also, Sal is calculating the probability that you would find more than one maple tree, it could be up to 10 trees.(3 votes)

- i still dont get it tho how do i do this without the explanation(3 votes)
- Assume there is a probability that an event will happen. You would not be interested if it would always happen as the probability then would be 1.

So, if it is less than 1, the counter probability must be greater than 0 therefore you have not only to take into account the one time, it might happen but also each other case in the row of events where it might happen.

As an intuition, a coin flip with a loaded die can be calculated for x throws, where the outcome of one throw is at least other, then expected.

Therefore you have to take into account these permutations/variations.(0 votes)

- Hi, I have a question about these "at least one" problems? At first, I didn't know how to do them so I calculated the probabilities of getting one bad chip, two bad chips, three bad chips, and all chips defective and added them together. However, I got 0.019216, which wasn't the answer. Why is that?(2 votes)
- I did some calculations based on common errors and found the likely error. The error is that you did not multiply the terms by the combinatorial coefficients

(4 choose 1)=4, (4 choose 2)=6, (4 choose 3)=4, and (4 choose 4)=1 for 1, 2, 3, and 4 bad chips, in this order.

You calculated

0.02*(0.98)^3 + (0.02)^2*(0.98)^2 + (0.02)^3*0.98 + (0.02)^4.

The correct calculation would be

4*0.02*(0.98)^3 + 6*(0.02)^2*(0.98)^2 + 4*(0.02)^3*0.98 + 1*(0.02)^4.

(Example for one of the terms: to find the probability that exactly 2 chips are bad, keep in mind that any 2 of the 4 chips can be bad, and so multiplication by (4 choose 2) = 6 is needed.)

Of course, it is much easier to just find 1 - P(no chips are bad) = 1 - (0.98)^4 than to go through all the calculations for 1, 2, 3, and 4 bad chips.

By the way, getting an answer like 0.019216 for this problem should immediately alert you that something's wrong, even before you submit the answer or look at the solution! Because the probability that a given chip is bad is 0.02, the probability that at least one of 4 chips is bad must clearly be larger than 0.02.

Have a blessed, wonderful day!(2 votes)

- Tossing a fair coin what is the probability of having 5 heads at the 10th toss?(2 votes)
- The number of possible outcomes after flipping 10 coins is 2¹⁰ = 1,024

The number of favorable outcomes is the number of ways we can arrange 5 Heads among 10 coins, which is

10!∕(5! ∙ (10 − 5)!) = 252

So, the probability of having exactly 5 Heads after flipping 10 coins is

252∕1,024 ≈ 0.2461(2 votes)

- Tell me if I have it right:

P(at least 1 x)= 1-P(not a single x)

P(at least 1 x)= 1-(P(not x) * P(not x) * P(not x)... for some number 'n' times)

P(at least 1 x)= 1-(P(not x)^n)*P(at least 1 x)= 1-((1-P(x))^n)*(2 votes) - In the explanation for 'surgical transplants' I don't understand 'in this situation it is easier...'. What is the alternative? And why is 'P ONE defective equal to 1- P 'all defective.'(1 vote)
- With 𝑃(𝑁) being the probability that exactly 𝑁 of the 8 patients reject their implants, then the probability that at least 1 patient rejects their implant is

𝑃(≥1) = 𝑃(1) + 𝑃(2) + ... + 𝑃(8)

However, the sum of*all*probabilities is 1, so

𝑃(0) + 𝑃(1) + ... + 𝑃(8) = 1 ⇒

⇒ 𝑃(1) + 𝑃(2) + ... + 𝑃(8) = 1 − 𝑃(0),

which is much easier to calculate.

– – –

Note that 𝑃(0) is the probability that none of the patients reject their implants, which is of course the same thing as the probability that all 8 patients accept their implant.(3 votes)

- How would I calculate probability for an event triggering at least two times, like for example if if there is .02 chance a virus will invade your computer upon a download and you downloaded 400 files, what is the probability that your computer would get at least two viruses?(2 votes)