If a pick three player (see Exercise 19) opts for \"box play, \" the player wins
ID: 3172441 • Letter: I
Question
If a pick three player (see Exercise 19) opts for "box play, " the player wins if his selection matches the winning number in any order For example, if he selected the number 157, he wins if the winning number is 715 or any other reordering of 157 How many different winning box play numbers are there if the three digits are different? That is, how many different ways are there to reorder a number such as 157? Find the probability of winning with box play if the three digits are different. How many different winning box play numbers are there if two digits are the same? That is, how many different ways are there to reorder a number such as 266? Find the probability of winning with box play if two digits are the same.Explanation / Answer
(a) There are 10 digits from 0-9. So the different arrangements that can be formed where all the digits are distinct is 10P3. i.e 10!/7!= 10x9x8 = 720 ways.
(b) We know that P(A)= Total number of favourable outcomes/Total number of outcomes. Here the total no. of outcomes is what we have found in (a) = 720. Lets take a look at our favourable outcomes. lets pick a random 3 digit distinct no say 572. Now to win i should get either of the following numbers 572,527,752,725,257,275 i,e 6 options (or 3! options...If it were 4 digits then 4! options).
Therefore my probability of winning = 6/720 = 1/120
(c) Here we are now saying that there are numbers where 2 digits are same, i.e 2 digits are repeated. which means we need to choose any 2 numbers from 0-9, (i.e 10C2, arrange them as 3 numbers which is 3! ways ie (10!/8!2!)x3!=270 ways. eg i pick the numbers 1&2, then the possible 3 digit numbers where 2 digits are repeating is 112,121,211,221,212&122.
(d) We know that P(A)= Total number of favourable outcomes/Total number of outcomes. Here the total no. of outcomes is what we have found in (a) = 270.
Lets take a look at our favourable outcomes. Suppose he picks a 3 digit number 266..then the different favourable options are 266,626 or 662 i.e for any number he picks there are only 3 favourable options.
Therefore my probability of winning = 3/270 = 1/90.
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