0
$\begingroup$

Hello can I ask how to approach this type of problem. A restaurant is giving away 3 different toys with each child's meal. What is the probability of getting all three toys when purchasing 5 meals?

$\endgroup$
  • $\begingroup$ I have a total oitcome of 3^5 equals to 243. I don't know the total possible outcome. The answer from the book is 0.6713 $\endgroup$ – Jan Navasca Oct 15 '19 at 9:32
5
$\begingroup$

A different approach is: There are $3^5$ different possibilities. With only toys A,B there are $2^5$ possibilities. Same for toys A,C or B,C. There are 3 possibilities with one toy.

Thus, using inclusion-exclusion: You get 3 toys in $3^5-3\times2^5+3=150.$

So the probability is $\frac{150}{243}=\frac{50}{81}$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thanks for the nice solution, Alberto. I edited your answer to use MathJax to make it look nicer. You can find a basic guide here: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Ingix Oct 15 '19 at 11:13
  • $\begingroup$ Thanks for your help. I am new here... $\endgroup$ – Alberto Saracco Oct 15 '19 at 11:17
  • 1
    $\begingroup$ Sure, I saw that (not sure if you can yourself see the "New contributor" button under your name). MathJax is derived from LaTex, so if you have experience with that, it shouldn't be a big problem. And using it makes longer solutions that much better to understand, especially if the formulas become more complicated. $\endgroup$ – Ingix Oct 15 '19 at 12:09
0
$\begingroup$

A tree of probability may be a good approach. Let us call the toys A,B,C.

Gift 1: You get toy A Gift 2: You either get again toy A (prob 1/3) or toy B (prob 2/3, you have AB)

Gift 3 in situation A (1/3): You either get again toy A (prob 1/3 - total 1/9) or toy B (prob 2/3 - total 2/9) Gift 3 in situation AB (2/3): You either get a repeat (prob 2/3 - total 4/9) or toy C (prob 1/3 - total 2/9) After gift 3: A - 1/9 AB - 6/9=2/3 ABC - 2/9

Gift 4 in situation A (1/9): You either get again toy A (prob 1/3 - total 1/27) or toy B (prob 2/3 - total 2/27) Gift 4 in situation AB (2/3): You either get a repeat (prob 2/3 - total 4/9) or toy C (prob 1/3 - total 2/9) After gift 4: A (impossible to get all 3) - 1/27 AB - 2/27+4/9=14/27 ABC - 4/9

Gift 5 in situation AB (14/27): you get toy C with prob 1/3: Total 14/81 You end with all 3 toys with probability 4/9+14/81=50/81 which is roughly 0,6172 (possibly you inverted two figures?)

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.