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Say we're talking about contraception, and the probability of one contraceptive, $A$, working is $99\%$, and the other, $B$ is also $99\%$. What is the probability of them working using both at the same time?

When I tried raising 99% to the power of 2, I realized I only got the effectiveness rate of contraception used at two separate events. Other similar operations haven't worked.

So I'm not sure what else to do here.

Edit: I'm aware medical statistics are extremely complicated, so I'm only assuming independent conditions for conventional purposes to get some idea of what the probability looks like.

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  • $\begingroup$ It's critical that we know the dependence between them, if any. That is to say, if $A$ fails then perhaps it failed for a reason and perhaps that reason will also lead $B$ to fail. $\endgroup$ – lulu May 21 at 23:43
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    $\begingroup$ @lulu (and OP): Or, alternatively, they use two rather different mechanisms that cover for each other, and the probability of at least one succeeding could be $1$. $\endgroup$ – Brian Tung May 21 at 23:54
  • $\begingroup$ @BrianTung exactly. Medical statistics are incredibly complicated precisely because of the various dependencies. $\endgroup$ – lulu May 21 at 23:55
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    $\begingroup$ Or, perhaps they're really two different brands of exactly the same drug. $\endgroup$ – BruceET May 22 at 4:56
  • $\begingroup$ @BruceET: Haha yeah. (I'm not sure that's a funny haha.) $\endgroup$ – Brian Tung May 22 at 14:33
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The method has worked if one of them worked or both worked. In short,

$$P(\text{method worked}) = 1-P(\text{both failed})$$

$$= 1-(0.01)^2$$

Edit:

I have assumed the two methods were independent when I said $P(\text{both failed}) = 0.01\times 0.01$

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    $\begingroup$ It's worth pointing out that this assumes independence between the effectiveness of the two measures. $\endgroup$ – Brian Tung May 21 at 23:47
  • $\begingroup$ Thanks Brian. Have edited my answer. $\endgroup$ – Vizag May 21 at 23:55
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By inclusion-exclusion, the probability that at least one measure works is equal to

\begin{align} P(\text{$A$ or $B$}) & = P(A) + P(B) - P(\text{$A$ and $B$}) \\ & = 1.98 - P(\text{$A$ and $B$}) \end{align}

If $A$ and $B$ are independent, then by definition,

$$ P(\text{$A$ and $B$}) = P(A)P(B) = 0.9801 $$

and then

$$ P(\text{$A$ or $B$}) = 1.98-0.9801 = 0.9999 $$

as in Vizag's answer. Otherwise, we need to account for dependence between $A$ and $B$, as represented in the foregoing expression by $P(\text{$A$ and $B$})$.


As is conventional in logical terminology, the use of "$A$ and $B$" and "$A$ or $B$" are logical connections. The conjunction "$A$ and $B$" means "$A$ is true and $B$ is true," while the disjunction "$A$ or $B$" means "$A$ is true or $B$ is true." In this case, the former does not mean "the physical combination of $A$ and $B$ works." (Note that it is difficult to come up with a meaningful interpretation of "the physical combination of $A$ or $B$.")

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    $\begingroup$ OP is asking "What is the probability of them working using both at the same time?" if we use P(A AND B)=0.98, this result is strange because each one by itself gives 0.99! That means using them together is not as good using them individually...Could it be so? $\endgroup$ – NoChance May 22 at 0:04
  • $\begingroup$ @NoChance I noticed the same thing! Intuition tells me the probability should be the same or slightly better, but we get worse probability assuming total independency. That's really strange. $\endgroup$ – Lex_i May 22 at 0:44
  • $\begingroup$ @Lex_i I think the above formula is applicable when the events are dependent. In your case they are independent, so the formula may be not suitable hence the strange result. What makes sense to me (not based on scientific formula), it should be 0.99 <= P <=1 $\endgroup$ – NoChance May 22 at 1:21
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    $\begingroup$ @NoChance: The term $A$ and $B$ is a logical conjunction—that is, "$A$ works and $B$ works"—not "the physical combination of $A$ and $B$ works." Similarly for $A$ or $B$: It means "$A$ works or $B$ works." $\endgroup$ – Brian Tung May 22 at 14:30
  • $\begingroup$ @Lex_i: See above comment to NoChance, and also the edit to my answer. $\endgroup$ – Brian Tung May 22 at 14:30

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