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I'm having a hard time understanding this probability thing right now.

I have exam next week and I'm not sure if I have time to doublecheck correct answers with my teacher before the exam. I made some effort on this but I'm not sure whether the logic is sound here.

Problem statement: The probability that a component will be acceptable (OK) is $0.92$. Three components are picked at random.

a) find prob that all three are acceptable

b) find prob that all three are failing

c) find prob that exactly two are acceptable and one is failing

d) find prob that at least two are acceptable

answers:

a] $0.92^3 = 0.778688$

b] $0.08^3=0.000512$

c] I would imagine that one way would be to sketch out the possible events such as for possible cases when you take three components. This is based on an earlier hazy recollection of mine. To be honest, I'm not sure why I sketched out these but it seemed like a good idea at the time xD. So, an $A$ would indicate acceptable component I suppose, and $F$ indicates failed component

  • $AAA$
  • $AAF$
  • $AFA$
  • $FAA$
  • $FFA$
  • $FAF$
  • $AFF$
  • $FFF$

strangely enough when I compute those letter-sequences as products such as $$AAA = P(\text{component acceptable})\cdot P(\text{component acceptable})\cdot P(\text{component acceptable})$$

Or $$FAA= P(\text{component Failure})\cdot P(\text{component Acceptable})\cdot P(\text{component Acceptable})$$ When I sum all the rows of three together I get the entire sample space $=1.0$

To my mind this would suggest that I have to sum together some of these rows such as $AAF+FAA+AFA$ are the cases when there is two acceptables and one failure I would imagine the probability is then $$P(\text{Two Acc} + \text{One failure})= 3\cdot (0.92^2 \cdot0.08)= 0.203136$$

Why do I have to sum those cases why can't I just have single calculation such as $0.92\cdot0.92\cdot0.08$ ? (the result is obviously three times less, but the method works for a-part and essentially also b-part)

d] when at least two are acceptable -> it means that ($3$ acc $+ 0$ fail) OR ($2$ acc $+1$ fail)

I would imagine that we would use the earlier result from c-part as help here. $$P(2\text{ Acc} +1\text{ Fail}) \text{ or } P(3\text{ Acc} +0\text{ Fail}) = 0.203136 + 0,92^3 = 0.981924$$

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  • $\begingroup$ Are you familiar with the binomial law? $\endgroup$ – krirkrirk Mar 25 '18 at 19:28
  • $\begingroup$ no sorry... I'm a bit shit in probability math. Also I'm confused about what exactly is the definition of "sample space", "event" ,and "outcome" and "experiment". Especially when we consider more complicated case than one single die-roll, what is the sample space when you roll 3-regular dice? $\endgroup$ – Late347 Mar 25 '18 at 22:12
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$(a)$ and $(b)$ are correct.

Why do I have to sum those cases why can't I just have single calculation such as $0.92\cdot0.92\cdot0.08$ ?

That is the probability of getting an acceptable component twice in a row followed by a failed component. However, we must account for the $3$ different ways to place a failed component, as your sample space correctly shows.

the result is obviously three times less, but the method works for a-part and essentially also b-part

That is because there is only one way to place $3$ successful components and similarly for placing $3$ failed components.

For $(c)$ summing the event space probabilities is one acceptable way of doing it. Namely

$$P(AAF)+P(AFA)+P(FAA)=3\cdot 0.92^2 \cdot 0.08\approx 0.203$$

You may also use the binomial distribution. The probability of $k$ successes in $n$ trials is given by

$$\begin{align*} P(X=k) &={n \choose k}p^k (1-p)^{n-k}\\\\ &={3 \choose 2}0.92^2\cdot 0.08\\\\ &\approx 0.203 \end{align*}$$

Again, for $(d)$ taking

$$P(AAA)+P(AAF)+P(AFA)+P(FAA)=\left(0.92^3\right)+\left(3\cdot 0.92^2 \cdot 0.08\right)\approx 0.982$$

is perfectly acceptable.

You could also do

$$P(X\geq 2)=\sum_{k=2}^3 {n \choose k}p^k (1-p)^{n-k}$$

and come to the same conclusion.

In R statistical software,

> sum(dbinom(2:3,3,.92))
[1] 0.981824

Note: Using the binomial distribution did not save us much (if any) time in this situation but you can imagine writing out all the possible scenarios for larger $n$ and $k$ would be quite tedious.

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  • $\begingroup$ ok, so I understood this correctly... We have a combined sample space which consists of outcomes. This sample space is denoted by the outcomes such as AAA, AAF, AFA, FAA ... Then we could have an Event with a probability such as Event E: exactly two acceptables and one failure. Then, we can say P(E) = P(AAF)+P(AFA) +P(FAA) $\endgroup$ – Late347 Mar 25 '18 at 22:31
  • $\begingroup$ Yes, the sample space is all of the possible outcomes. The event space is all the possible outcomes that satisfy our criteria. $\endgroup$ – Remy Mar 25 '18 at 22:34
  • $\begingroup$ So, If I wanted to define Event F: first pick acceptable, second pick acceptable, third pick failure. Then, P(F) = 0,92^2 * 0,08 , and there is no need to add, if we were only interested in that particular event? I realize that this event might not be the same as was originally asked for but is still still essentially correct reasoning? $\endgroup$ – Late347 Mar 25 '18 at 22:39
  • $\begingroup$ Yes that would be correct. $\endgroup$ – Remy Mar 25 '18 at 22:42
  • $\begingroup$ Ok, so typically if I have three times picking something randomly, then I presume that the singular outcome in this case would be AAA, and then the trial (3 picked items) is concluded. Other possible outcomes are AAF, AFA,FAA etc... ??? $\endgroup$ – Late347 Mar 25 '18 at 22:52

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