QR problems!

[sfoksn]

---

Members don't see this ad.

1) HOW MANY DIFFERENT WAYS CAN YOU PUT ON A HAT, TIE, AND THEN A PAIR OF SHOES, IF YOU HAVE 10 DIFFERENT TIES, 20 DIFFERENT PAIRS OF SHOES, AND 15 DIFFERENT HATS?
A. 3000
B. 300
C. 45
D. 1500
E. 150

2) WHAT IS THE PROBABILITY THAT YOU WILL THROW A SUM OF 6 IN TWO TOSSES OF A DIE?
A. 1/18
B. 1/12
C. 1/9
D. 5/36
E. 1/6

I will post answer soon! Thank you for helping me.

 

[herkulease]

1. 3000 but it just feels like wrong for some reason.
2. 1/12

 

[dentalWorks]

1) HOW MANY DIFFERENT WAYS CAN YOU PUT ON A HAT, TIE, AND THEN A PAIR OF SHOES, IF YOU HAVE 10 DIFFERENT TIES, 20 DIFFERENT PAIRS OF SHOES, AND 15 DIFFERENT HATS?
A. 3000
B. 300
C. 45
D. 1500
E. 150

2) WHAT IS THE PROBABILITY THAT YOU WILL THROW A SUM OF 6 IN TWO TOSSES OF A DIE?
A. 1/18
B. 1/12
C. 1/9
D. 5/36
E. 1/6

I will post answer soon! Thank you for helping me.

1) 3000
2) this is "OR":
1 + 5 ==> P of 1 is 1/6, P of 5 is 1/6 so... 1/6 * 1/6 = 1/36
OR
2 + 4 ==> P of this happening is 1/36
OR
3 + 3 ==> P of this happening is 1/36 also
OR
4 + 2 ==> P of this happening is 1/36 also
OR
1 + 5 ==> P of this happening is 1/36 also

so answer = 5 (1/36) = 5/36

 

[sfoksn]

Hello Herkulease, thank you for your answer.

I got the same answer for 1, but the solution says B is the answer.

For the second one, would you mind telling me how you got 1/12? i got 5/36.

 

[herkulease]

I got 1/12 because there are 3 combination of getting 6

1,5
2,4
3,3

I assumed 5,1 and 4,2 were the same as the reverse. If I counted them as separate events then yes I'd get 5/36 also.

With number 1 I"m not sure how they get 300. I'll have to think about that.

 

[NewBee12]

1) 10x20x15 = 3000
2) 5/36

 

[preppydoc]

I got 1/12 as well, which is actually correct? Is it 5/36 or 1/12? So sorry, butI wonder what the testers would think would be correct?

 

[UCB05]

If you count 36 different 2-die combinations, you're counting all landings of the first throw 1-6 times the second throw, 1-6. If you're then counting the number of ways the die tosses add up to 6, then you have to take into consideration all tosses of one die independently paired with the other, i.e 1-5 is different from 5-1. There should be 5 such pairs of tosses, 1-5, 2-4, 3-3, 4-2, 5-1, so I'd go with 5/36 too.

sum of:
2: 1/36 (1-1)
3: 2/36 (1-2, 2-1)
4: 3/36 (1-3, 2-2, 3-1)
5: 4/36 (1-4, 2-3, 3-2, 4-1)
6: 5/36 (1-5, 2-4, 3-3, 4-2, 5-1)
7: 6/36 (1-6, 2-5, 3-4, 4-3, 5-2, 6-1)
8: 5/36 (2-6, 3-5, 4-4, 5-3, 6-2)
9: 4/36 (3-6, 4-5, 5-4, 6-3)
10: 3/36 (4-6, 5-5, 6-4)
11: 2/36 (5-6, 6-5)
12: 1/36 (6-6)

total probability: 36/36

 

[Streetwolf]

It's 5/36.