Can someone explain these concepts in probability?

  • Picking an object with replacement / without replacement.
  • Picking a few objects --- all at the same time.



2 Answers 2


You are at a store. There's a 50% sale resulting in a copious amount of people. You are not able to look at the dresses kept for sale in a huge basket—you manage to get your arm between two hips and take out a dress. You recall that the basket contains 4 yellow and 4 blue dresses.

With replacement: You take out a yellow dress (and you wanted the blue dress). You keep the yellow dress back in the basket in disappointment. You calculate the theoretical chances of getting a blue dress this time. Here's your work:

After I put back the yellow dress, I knew that then, the basket now contains 4 yellow and 4 blue dresses.

There are 4 dresses—the blue ones—out of 8, that will please me. Oooh. I know that formula!$$\rm P(event) = {\text{number of desired outcomes }\over \text{total number of possible outcomes}}= {4 \over 8}= {1 \over 2}$$

Without replacement: You take out a blue dress which brings a smile on your face. “Yay,” you exclaim. Oops... you want one more for your sister. You don't put the blue dress back in the basket, but you try your luck for another one. You calculate the theoretical chances of getting a blue dress again. Here goes your work:

Okay... I know that now, the basket contains 3 blue and 4 yellow dresses because a blue dress is in my hand. I'd love if I get one of those 3 blue dresses and now there are 7 dresses in total. You apply the formula again. $$P(event) = {\text{number of desired outcomes }\over \text{total number of possible outcomes}} = {3\over 7}$$

A few objects... at the same time: You're sad because you got a yellow dress. You throw that yellow dress furiously back into the basket. Now, you try your luck with multiple dresses at the same time. You decide that you will pick 3 at the same time and you try to calculate how luck favours you now. Here it is:

Okay... this is hard but I can do it. There are 3 blues and 4 yellows. Hmm, but what if I think of those three dress something like different objects? So even if I pick three object at once, I will start with one dress then move on to the other... then the next. So the probability that the first dress is blue will be ${3/7}$... then I have only 6 dresses left and 2 blue ones left assuming I pick a blue one which makes the probability $2/6 $. Then the next is $1/5$. I'd multiply these three individual events together to get the probability of picking 3 bluer dresses at once. I get this:$${3 \over 7}\times{2\over 6}\times{1 \over 5} = {2 \over 70}$$ Eh! It's really hard to make it happen.

"Replacement" is really the thing you do when you take out an object from some collection and put it back. "Not replacing" is just the opposite: you don't put it back - you keep it. The third concept is self-described.

  • $\begingroup$ Just to make sure I understood correctly , for example If we Have 10 cards numbered from 1 to 10 and we want to pick 3 of them at the same time What will be be the total number of cases ? C(10,3) $\endgroup$
    – Node.JS
    Nov 17, 2012 at 18:50
  • $\begingroup$ Yes. That's right! $\endgroup$
    – P.K.
    Nov 17, 2012 at 20:03

Imagine that there is a room of 20 people. They are sitting and listening to me sing. Consider that I have to pick 3 of them to come and sing with me on stage. I choose them by randomly calling out a number from 1 to 20. Now, which people come sing with me? There are many ways I can do this:

  • I shout out a number between 1 and 20 and the corresponding person comes sings with me. After that, I ask him to go back to his seat. Then I shout out another number. If I allow myself to shout out a number which I have already shouted, it is with replacement. If on the other hand, I prevent people who have already sung to not sing again, it is without replacement.
  • Instead of having them come over one at a time, I can simultaneously shout 3 numbers and have all three come up on stage at the same time. That is picking all at the same time.

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