Usain Bolt, Tyson Gay, Andre De Grasse, Roger Federer, and Fat Joe will compete

Question

Usain Bolt, Tyson Gay, Andre De Grasse, Roger Federer, and Fat Joe will compete in the 100 metres at the 2020 Olympics in Tokyo. The runners can finish in any possible order, except that Fat Joe can never beat Usain Bolt. We assume that ties are not possible. How many ways are there for the race to finish? Usain Bolt has three 10-sided dice. Fat Joe has two 10-sided dice, Usain Bolt rolls his dice 100 times, and counts how many times he got a total of 14. Then Fat Joe rolls his two dice 100 times, and he counts how many times he got a total of 14. Whoever got 14 more times wins. Who is more likely to win? Rafael Nadal selects a tennis ball from an ura containing 1000 tennis balls. The tennis balls are numbered from 1 to 1000. What is the probability that the number on the selected tennis ball is divisible by 2, 7 or 31? Fat Joe has a biased 6-sided die denoted by D1: the probability of rolling number i with D1 is 21i-1/36. Fat Joe also has an unbiased die denoted by D2. He rolls the two dice, and he tells you that the total is less then 7. (You don't see the dice.) What is the probability that D2 is 3?

Explanation / Answer

1. Bolt, Tyson, Andre, Roger, Joe compete in 100m. Assuming there are no ties, then the number of possible orders can be determined as follows:

No. of choices for 1st Place x No. of choices for 2nd place x.... x No. of choices for 5th place

= 5 x 4 x 3 x 2 x 1 = 120 total possible ways are there for the five people to finish the race.

Now the condition is given that Joe can never beat Bolt. That means, Joe will never be the winner of the race (or never be 1st). Then the choices for places changes as follows:

For 5 places, Bolt has first 4 places to choose from and depending on the choice of Bolt Joe chooses. If bolt chooses first, then Joe has 4 choices, If Bolt chooses second Joe has 3 choices and so on. No. of ways to arrange Bolt and Joe are 4+3+2+1 = 10. For the remaining 3 spots, number of ways the remaining can choose are 3x2x1 = 6. Thus, Total number of ways = 10*6 = 60 ways

2. Ways of getting a 14 from 3 10-sided die are (10, 1, 3), (10, 2, 2), (9, 3, 2), (9, 1, 4), (8, 1, 5), (8, 2, 4), (8, 3, 3), (7, 1, 6), (7, 2, 5), (7, 3, 4) = 10 ways but each of the 3 outcomes of a die can be in any order = 3 x 2 x 1 = 6

Therefore, total no of ways = 10 x 6 = 60

Now notice, that (10, 2, 2) and (8, 3, 3) are the only ones where the number of permutations will be 3 instead of 6 because 2 of the die have the same outcome. Thus, we subtract 6 outcomes from the total 60 outcomes.

Thus, total no. of ways to have 14 = 60 - 6 = 54

Similarly for 2 10-sided die, (10, 4), (9, 5), (8, 6), (7, 7) = 4 ways to come up 14. And flipping the outcomes of the 1st and 2nd die (Note: (7,7) would not flip), we have 3 outcomes as (4, 10), (5, 9) and (6, 8). Therefore, total no. of ways to get 14 = 7.

Probability for getting 14 for 3 die case = 54/(103) = 0.054

Probability for getting 14 for 2 die case = 7/(102) = 0.07

Thus, Joe is more likely to win.

3. D(2): No. of balls numbered with even numbers (divisible by 2) = 1000/2 = 500

D(7): No. of balls divisible by 7 = 1000/7 = 142 (ignoring the remainder, as numbers will be <= 1000 )

D(31) :No. of balls divisible by 31 = 1000/31 = 32

Now, there are numbers between 1 - 1000 that are divisible by any two or the given three numbers (2, 7, 31) or can be divible by all the three numbers, we take out those number of cases.

D(2, 7): No of ball divisible by 2,and 7 = 1000/(2x7) = 71

D(2, 31) = 1000/(2x31) = 16

D(7, 31) = 1000/(7x31) = 4

D(2, 7, 31) = 2

Therefore Answer = D(2) + D(7) + D(31) -D(2,7) - D(2,31) - D(7,31) + D(2,7,31) = 500+142+32-71-16-4+2 = 585

4. Prob. for the biased coin D1 for outcomes 1 to 6 are 1/36, 3/36, 5/36 ... 11/36 (replacing in formula)

Prob. for the unbiased coin D2 for outcomes 1 to 6 are 1/6 for each.

Total is <7. The outcomes for which we get <7 are (1,1), (1, 2),... (1,5), (2, 1), (2, 2) and so on. We can draw a 6x6 matrix to check the outcomes. No. of such outcomes = 15

Now the sample space becomes these 15 outcomes.

No. of outcomes for D2 being 3 given that result is <7) = 3 outcomes. P(D2 being 3/ result 7) = 3/15

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