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
Solution 1:
"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.
Solution 2:
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
and then added terms like
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.