Celebrity pics maths Qn
Pop quiz! Slightly modified details about the question as I cannot remember exactly. =/
"A petrol station gives its motorists a card with a celebrity pic on it whenever a motorist tops up petrol. [Sounds awesome!] There are 10 kinds of different pics, and 1 of the 10 pics is given randomly for every top-up. "
*Assumption: Equal probability, unlimited stock, etc. XD*
So this random motorist went to this petrol station and topped up 4 times, and hey he got 4 pics as thank you gifts. ^^ Here comes the questions~!
So for the 4 pics the motorist received...
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Q: Let's name 2 of the pics as X and Y. What's the probability that the motorist would get at least X, Y or both?
A: 0.5904
Solution
Calculate the number of ways the motorist would get 4 pics without X or Y. Number of ways = 8 x 8 x 8 x 8 = 4096
Total number of ways he could get 4 pics without restrictions = 10 x 10 x 10 x 10 = 10,000
Thus, the number of ways the motorist could get X, Y or both = 10000 - 4096 = 5904
Probability = 5904/10000 = 0.5904 ^^
--
Q: What's the probability that the motorist would get exactly 3 different pics?
A: 0.432
Here are some of the suggested (wrong) workings, for entertainment purposes. XD The wrong parts are highlighted in red.
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By Jeff:
_ X _ Y _
X and Y are 2 different pics, so there will be 10 ways to choose X, 9 ways to choose Y. Permutate them, number of ways = 10 x 9 x 2 = 180.
_ indicates the places which you can put Z in, where Z is the repeated pic. Since there are 8 ways to choose Z, and 3C2 ways to put 2 Z in the 2 _, therefore number of ways to put 2 Z = 8 x 3C2 = 24.
Combining both, number of ways = 180 x 24 = 4320.
Probability = 4320 / 10000 = 0.432. [Surprisingly correct answer, wrong concept]
WL: Eh! Why the 2Z cannot be together!? ZZ sia!
Jeff: Then I realised I dunno why there's a need to x2 for X and Y.
--
By WL:
Since there are 10 ways to choose X, 9 ways to choose Y, 8 ways to choose Z. Probability = 1/10 + 1/9 + 1/8 = 0.336 (3 sig)
Jeff: How the hell did u even get this? Why plus?
WL: Ok la, the english is there, even though the maths isn't there, 0.336 and 0.432, close enough right? :D
Jeff: I don't understand both your english and maths sia!
WL: To understand better, I rephrased the question to, 'What is the probability of X, Y or Z kena repeated?'
Jeff:
Solution
There are 10 ways to choose X, 9 ways to choose Y, 8 ways to choose Z. Z is gonna be the repeated one! :D
Case 1 - The 2 Z are seperated
_ X _ Y _
_ indicates the places which you can put Z in, where Z is the repeated pic. There are 3C2 ways to put 2 Z in the 2 _, thus total number of ways = 3 x [10 x 9 x 8] = 2160. Including the number choices.
Case 2 - The 2 Z is counted as a group.
_ X _ Y _
_ indicates the places which you can put ZZ in, where Z is the repeated pic. There are 3C1 ways to put the group in the _, thus total number of ways = 3 x [10 x 9 x 8] = 2160. Including the number choices.
Total number of ways = 2160 + 2160 = 4320
Probability = 4320 / 10000 = 0.432 ^^
======================
All right, that's all! Somehow I feel it's easier to visualise this problem as a 4D thing instead of the pics. Agree? XD
*Assumption: Equal probability, unlimited stock, etc. XD*
So this random motorist went to this petrol station and topped up 4 times, and hey he got 4 pics as thank you gifts. ^^ Here comes the questions~!
So for the 4 pics the motorist received...
--
Q: Let's name 2 of the pics as X and Y. What's the probability that the motorist would get at least X, Y or both?
A: 0.5904
Solution
Calculate the number of ways the motorist would get 4 pics without X or Y. Number of ways = 8 x 8 x 8 x 8 = 4096
Total number of ways he could get 4 pics without restrictions = 10 x 10 x 10 x 10 = 10,000
Thus, the number of ways the motorist could get X, Y or both = 10000 - 4096 = 5904
Probability = 5904/10000 = 0.5904 ^^
--
Q: What's the probability that the motorist would get exactly 3 different pics?
A: 0.432
Here are some of the suggested (wrong) workings, for entertainment purposes. XD The wrong parts are highlighted in red.
--
By Jeff:
_ X _ Y _
X and Y are 2 different pics, so there will be 10 ways to choose X, 9 ways to choose Y. Permutate them, number of ways = 10 x 9 x 2 = 180.
_ indicates the places which you can put Z in, where Z is the repeated pic. Since there are 8 ways to choose Z, and 3C2 ways to put 2 Z in the 2 _, therefore number of ways to put 2 Z = 8 x 3C2 = 24.
Combining both, number of ways = 180 x 24 = 4320.
Probability = 4320 / 10000 = 0.432. [Surprisingly correct answer, wrong concept]
WL: Eh! Why the 2Z cannot be together!? ZZ sia!
Jeff: Then I realised I dunno why there's a need to x2 for X and Y.
--
By WL:
Since there are 10 ways to choose X, 9 ways to choose Y, 8 ways to choose Z. Probability = 1/10 + 1/9 + 1/8 = 0.336 (3 sig)
Jeff: How the hell did u even get this? Why plus?
WL: Ok la, the english is there, even though the maths isn't there, 0.336 and 0.432, close enough right? :D
Jeff: I don't understand both your english and maths sia!
WL: To understand better, I rephrased the question to, 'What is the probability of X, Y or Z kena repeated?'
Jeff:
Solution
There are 10 ways to choose X, 9 ways to choose Y, 8 ways to choose Z. Z is gonna be the repeated one! :D
Case 1 - The 2 Z are seperated
_ X _ Y _
_ indicates the places which you can put Z in, where Z is the repeated pic. There are 3C2 ways to put 2 Z in the 2 _, thus total number of ways = 3 x [10 x 9 x 8] = 2160. Including the number choices.
Case 2 - The 2 Z is counted as a group.
_ X _ Y _
_ indicates the places which you can put ZZ in, where Z is the repeated pic. There are 3C1 ways to put the group in the _, thus total number of ways = 3 x [10 x 9 x 8] = 2160. Including the number choices.
Total number of ways = 2160 + 2160 = 4320
Probability = 4320 / 10000 = 0.432 ^^
======================
All right, that's all! Somehow I feel it's easier to visualise this problem as a 4D thing instead of the pics. Agree? XD
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