# Probability of receiving gifts on time

Alex plans to order a birthday gift for his friend from an online retailer. However, the birthday coincides with the festival season during which there is a huge demand for buying online goods and hence deliveries are often delayed. He estimates that the probability of receiving the gift, in time, from the retailers A, B, C and D would be 0.6, 0.8, 0.9 and 0.5 respectively.

Playing safe, he orders from all four retailers simultaneously. What would be the probability that his friend would receive the gift in time?

My solution:-

If we even get 1 gift on time, the task would be done, so I thought of finding 1-P(No gift on time)

P(No gift on time)=P(A)*P(Not on time | A) + P(B)*P(Not on time | B) + P(C)*P(Not on time | C) + P(D)*P(Not on time | D)

Choosing any of gifts among A,B,C,D are equally likely therefore P(A)=P(B)=P(C)=P(D)=1/4

P(No gift on time) = 1/4 * (0.4 + 0.2 + 0.1 + 0.5) = 0.3

Therefore P(At least 1 gift on time)= 1-0.3 = 0.7

but the answer is given as 0.996 , what mistake am I making ?

Update :-

I have understood that P(A)=P(B)=P(C)=P(D)=1 as it is certain that alex has ordered the gifts from all these retailers, but my question now is why is the following notation wrong ?

P(No gift on time)=P(A)*P(Not on time | A) + P(B)*P(Not on time | B) + P(C)*P(Not on time | C) + P(D)*P(Not on time | D)

A= event of ordering gift from retailer A

P(Not on time | A) = Probability of gift not arriving on time given that it was ordered from retailer A

• Please do use MathJax. Commented Dec 29, 2021 at 21:31

"f we even get 1 gift on time, the task would be done, so I thought of finding 1-P(No gift on time)"

This is correct!

Now the question is how to compute P(no gift on time). Note that you order from all four, so $$P(A) = P(B) = P(C) = P(D) = 1$$, not 1/4.

Finally: "no gift on time" means "A not on time AND B not on time AND C not on time AND D not on time". AND, not OR. I think you can take it from there.

• hi, thanks a lot for the answer, can you please elaborate why are we not using P(A)=P(B)=P(C)=P(D)=1/4 and rather using 1, how would the question be framed then for us to use probability of 1/4 ? Commented Dec 27, 2021 at 18:36
• so, will the we do like this P(No gift on time)=P(A)*P(Not on time | A) * P(B)*P(Not on time | B) * P(C)*P(Not on time | C) * P(D)*P(Not on time | D) ? Commented Dec 27, 2021 at 18:37
• Yes. P(A) is the probability of getting a gift from shop A (at all). Since Alex ordered the gift from shop A we can be confident that it will arive sooner or later, so P(A) = 1. Similar P(B) = 1 since he ALSO ordered at B. To get at P(A) = P(B) = P(C) = P(D) = 1/4 the question would probably involve Alex ordering at only one store (A, B, C OR D) which he moreover would choose at random. Commented Dec 27, 2021 at 19:48
• It is true that Alex takes his 'better save than sorry'-approach a bit far by ordering four identical copies of the gift, even if he wants to give away only one and put the other three in the trash (okay the problem doesn't specify that. Maybe he keeps them or gives them to other friends). However, the alternative version where he orders only one but randomly picks the shop is ALSO not very realistic. If he were to order at only one shop, he would logically order from only shop C because that has the highest probability of delivering in time. Commented Dec 27, 2021 at 19:50
• Your notation P(Not on time | A) is a bit confusing. To me it sounds like "the probabiilty of the gift arriving on time given that the gift comes from A". But what is "the" gift here? There are four gifts, one from A, one from B, one from C and one from D. Each of which can be on time or too late, leaving sixteen possible outcomes. One of those 16 is really undesirable (all four are too late) and we want the probability of that happening (or rather not happening). Commented Dec 27, 2021 at 19:53

There are several answers already that do not seem to convince you, so let me try a rephrasing of a similar problem.

Suppose you place two bets. One is on a coin flip and you win if it's heads. The other is on a die roll and you win if it's a $$6$$. What is the probability that you win at least one bet?

There are 12 possible outcomes, like (head, $$5$$), (head, $$6$$), (tail, $$3$$) and so on. You win at least one of the bets with (head, anything) and with (tail, $$6$$). That accounts for $$7/12$$ of the possible outcomes, so that is the probability. The fact that you win both bets with (head, $$6$$) does not matter.

To work this out another way, look at how you might lose. You lose the first bet with probability $$1/2$$ and the second with probability $$5/6$$. Since those are independent, you lose both with probability $$(1/2)\times(5/6) = 5/12$$ so the probability that you win at least one is the complementary $$7/12$$.

The reasoning in your problem is the same. The numbers are harder.

You ask explicitly what is wrong with

P(Not on time | A) = Probability of gift not arriving on time given that it was ordered from retailer A

P(A)* P(Not on time | A)

That is the answer to a different question. Suppose (in my example) you don't place two bets, you place just one. And suppose you decide which one (coin or die) by flipping another coin first. Then the probability that you bet on the (second) coin is $$1/2$$. So is the probability that you bet on the die. Then you need conditional probability to find that probability that you win the whole game is: $$\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{6} = \frac{1}{3}.$$ It's no surprise that the answer is in between the probabilities of winning each of the individual bets. In fact it's just halfway between them.

What you did in your solution was to answer the gift question if Alex ordered from just one retailer, and chose which one at random with equal probability ($$1/4$$ for each). You got the correct answer, which is the average of the four probabilities of success in each.

• Thank you so much @Ethan Bolker, that clarified a lot of doubts for me, but I am stuck at your last example Your argument would be correct if you placed just one bet (not two) and chose the one to place with the flip of another coin. Then you would lose the single bet with probability , can you elaborate this a bit so that I can get a complete understanding, I am not sure which probability are we calculating in the last example of yours, thanks again :) Commented Dec 29, 2021 at 23:33
• @Fin27 I have edited the answer. Hope it helps. Commented Dec 30, 2021 at 1:16

For no gift to arrive on time, each one must not individually. So instead of adding probabilities, you must multiply them. This gives $$0.4\times0.2\times0.1\times0.5 = 0.004$$ The complement of which is the answer.

In your answer it was as if you were ordering from one company at random.

• did you find probability of P(no gift to arrive on time) = P(A)*P(Not on time | A) * P(B)*P(Not on time | B) * P(C)*P(Not on time | C) * P(D)*P(Not on time | D) ? what is your reasoning of taking P(A)=P(B)=P(C)=P(D)=1 ? Commented Dec 27, 2021 at 18:39
• What are events A, B, C and D here? If you express them in words you may find it easier to see why this is not the right way of thinking about the question. Commented Dec 27, 2021 at 22:24
• A=ordering gift from shop A, B=ordering gift from shop B,.... and by P(Not on time|A) would mean the probability of gift not arriving on time given that it was ordered from shop A = (1-0.6)=0.4 , is there any problem with this notation ? please let me know Commented Dec 28, 2021 at 22:55

why is the following notation wrong ?

P(No gift on time)=P(A)*P(Not on time | A) + P(B)*P(Not on time | B) + P(C)*P(Not on time | C) + P(D)*P(Not on time | D)

In conditional probability, $$P(A\mid B)$$ is the probability of $$A$$ occuring, assuming $$B$$ is known to have occurred. Implicit in this definition is that $$A$$ and $$B$$ are seperate, independantly defined events.

When you write $$P(Not\ on\ time\mid A)$$, it seems that you mean "the probability that the gift ordered from A arrives late". This definition doesn't really fit the "probability of X given Y" format of conditional probability. Even if we add a "given that the gift was ordered from A", this would help the second half, but the problem remains that the event $$Not\ on\ time$$ is not one event but four, corresponding to the four gifts ordered.

But the most critical issue is that you seem to have used the formula

$$P(A)=P(B_1)P(A\mid B_1)+P(B_2)P(A\mid B_2)+\cdots+P(B_k)P(A\mid B_k)$$

without any regard to what this equation means and how it is meant to be used. To use this formula, $$A$$ may be any event, but $$\{B_i\}$$ must be a partition of the event space. (That is, each possible outcome must satisfy exactly oneof the events in $$\{B_i\}$$.) What you've done looks similar, but critically $$A,B,C,D$$ are not exclusive--every outcome satisfies all four.

Interestingly, the formula you gave is the answer to a different but related problem. Suppose instead of a safe shopper, you instead had a lazy shopper who ordered a single gift from a random supplier (perhaps the first entry in a non-deterministoc search engine). Then you have that the events $$A,B,C,D$$ constitute a partition of the event space. (Gifts were ordered from exactly one of the suppliers.) Then, letting $$L$$ be the event the present arrives late, we have

$$P(L)=P(A)P(L\mid A)+P(B)P(L\mid B)+P(C)P(L\mid C)+P(D)P(L\mid D)$$

where $$P(L\mid A)$$, etc. are the probabilities of the gift arriving late from a given supplier, and $$P(A)$$, etc. are the probabilities of choosing a particular supplier to begin with.

Hopefully this helps you understand the use case of this conditional probability equation, and why the problem you gave does not fall into it. If this does not particularly address your specific issue, please leave a comment to help me to better understand where you're at.

Pr(receive from any store) = 1 - Pr(not receive from any store) = 1 - 0.4* 0.2 * 0.1*0.5 = 0.996

The reason for the incorrectness of $$P(\text{No gift on time})=P(A)*P(\text{Not on time} | A) + P(B)*P(\text{Not on time} | B) + P(C)*P(\text{Not on time} | C) + P(D)*P(\text{Not on time} | D)$$ is that the law of total probability you used above, works only for distinct events and note that $$\text{Not on time} | A$$ and $$\text{Not on time} | B$$ are NOT distinct for the following reason:

$$\text{Not on time} | X$$ means that an ordering from $$X$$ is delivered with delay. Based on this, $$[\text{Not on time} | A]\cap [\text{Not on time} | B]$$ means that both $$A$$ and $$B$$ cannot deliver the gift on time, which is indeed possible.

Although it is possible to correct the above formula by adding probabilities of intersections, what you need to do is to start solving the problem with a different approach. Note that $$\text{Not on time} | X$$ and $$\text{Not on time} | Y$$ are independent, not distinct and to use this property we can write $$P(\text{[No gift on time|A]\cap [No gift on time|B]\cap [No gift on time|C]\cap [No gift on time|D]}){=1-P(\text{Not on time} | A)P(\text{Not on time} | B)P(\text{Not on time} | C)P(\text{Not on time} | D). }$$ Hence $$P{=1-(1-0.6)(1-0.8)(1-0.9)(1-0.5) \\= 1-(0.4)(0.2)(0.1)(0.5) \\= 1-0.004=0.996. }$$