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Probabilities involving "at least one" success

Example 1 : Defective chips

A manufacturer of processing chips knows that 2% 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 1: Problem A
Find the probability that a randomly selected chip is NOT defective.
P(defective)=0.02
P(not defective)=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Example 1: Problem B
Find the probability that all 4 chips are NOT defective.
Round your answer to three decimal places.
P(all 4 NOT defective)=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Example 1: Problem C
Find the probability that at least one of the selected chips is defective.
Round your answer to three decimal places.
P(at least one defective)=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

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%. The rest of the patients successfully accept the implant.
Assume that the results for each patient are independent.
Example 2
In a random sample of 8 of these surgeries, find the probability that at least one patient rejects the implant.
Round your answer to three decimal places.
P(at least one rejects)=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Example 3: Free-throws

Esther is a basketball player who makes 75% of the free-throws she attempts. Assume that the results of each shot are independent.
Example 3
If Esther attempts 3 free-throws, find the probability that she misses at least one free-throw.
Round your answer to three decimal places.
P(at least one miss)=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Generalizing the strategy

In general, we can use these strategies:
P(at least 1 success)=1P(all failures)
or similarly,
P(at least 1 failure)=1P(all successes)

Want to join the conversation?

  • leaf grey style avatar for user sumit kothari
    So, P(at least 1 success)=1−P(all failures)

    then what will be P(at least 2 success) ?
    (20 votes)
    Default Khan Academy avatar avatar for user
    • primosaur seed style avatar for user Ian Pulizzotto
      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!
      (60 votes)
  • leaf green style avatar for user Oliver
    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)
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    • leaf orange style avatar for user Yevgeny Kozlov
      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
      (28 votes)
  • blobby green style avatar for user vigneshboserajan
    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)
    Default Khan Academy avatar avatar for user
    • primosaur seed style avatar for user Ian Pulizzotto
      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).
      (18 votes)
  • duskpin ultimate style avatar for user BeeGee
    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)
    Default Khan Academy avatar avatar for user
    • leaf grey style avatar for user dashpointdash
      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)
  • primosaur seedling style avatar for user Darknessas_Dark27
    i still dont get it tho how do i do this without the explanation
    (3 votes)
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    • leaf grey style avatar for user dashpointdash
      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.
      (1 vote)
  • duskpin ultimate style avatar for user sisipanda
    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)
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    • primosaur seed style avatar for user Ian Pulizzotto
      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)
  • leaf green style avatar for user Specialrelativityhardbuticandoit
    Why can't I use the probability that something is not going to happen and then times it, instead of taking it away from one. To illustrate my point further -
    (given practice problem)
    Surgeries involving implants sometimes result in the patient's body rejecting the implant. A certain surgery has a rejection rate of 11%. The rest of the patients successfully accept the implant.
    Assume that the results for each patient are independent.
    In a random sample of 8 of these surgeries, find the probability that at least one patient rejects the implant.

    to solve this problem the way we are given the formulas it is
    1 - (0.89)^8 = 0.606

    but why can it solve it as 0.11^8 as that's the rejection rate and I don't have to take it away from 1 later on yet the answer is much different. Could anyone please tell me the reason?
    (2 votes)
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    • blobby green style avatar for user daniella
      Using the rejection rate (11%) raised to the power of 8 (0.11^8) would give you the probability that all eight surgeries result in rejection, not the probability of at least one rejection. To find the probability of at least one rejection, you need to consider the complement of the event where all surgeries are successful (i.e., no rejection). So, you calculate the probability of no rejection (0.89^8) and subtract it from 1 to find the probability of at least one rejection.
      (2 votes)
  • blobby green style avatar for user Rogerio Almeida
    Tossing a fair coin what is the probability of having 5 heads at the 10th toss?
    (2 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      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)
  • blobby green style avatar for user mikaylab24
    how do we know wheter to use the percent given or subtract it from 1?
    (2 votes)
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    • blobby green style avatar for user daniella
      Whether to use the given percent or subtract it from 1 depends on the situation and what you're trying to find. If you're asked to find the probability of an event happening (e.g., at least one patient rejects the implant), you can use the complement rule by subtracting the probability of the event not happening from 1.
      (2 votes)
  • male robot johnny style avatar for user agguram
    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)
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