Math Questions... (Probablitity)

[frogger33]

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Hey guys.. There are two probablity questions that I cannot figure out. Can someone PLEASE help!! Thanks guys. I am really starting to stress since I am retaking the DAT at the end of the month and I have not even started anything but math yet....

1. What is the probability of getting 6 tails out of nine tosses of a fair coin??

a) 21/128
b) 21/64
c) 84/128
d) 22/64
e) 42/64

The correct choice is A

2. What are your chances of rolling a seven with 2 dice in one toss?

a) 1/6
b) 1/3
c) 1/36
d) 3/36
e) 1/2

The correct choice is A

 

[g3k]

Hey guys.. There are two probablity questions that I cannot figure out. Can someone PLEASE help!! Thanks guys. I am really starting to stress since I am retaking the DAT at the end of the month and I have not even started anything but math yet....

1. What is the probability of getting 6 tails out of nine tosses of a fair coin??

a) 21/128
b) 21/64
c) 84/128
d) 22/64
e) 42/64

The correct choice is A

2. What are your chances of rolling a seven with 2 dice in one toss?

a) 1/6
b) 1/3
c) 1/36
d) 3/36
e) 1/2

The correct choice is A

I can explain the second one... THere are a total of 6 sides on each dice so a total of 36 combinations. One can get 6 combinations of 7 on the dice ... 6,1 :2,5:3,4:4,3 :5,2 :6,1 .. so 6/36 =1/6

 

[crazy_sherm]

This link should help you with your first question. Check the equation under number 5.

http://www.saliu.com/theory-of-probability.html

I don't remember anything quite this complicated showing up on the actual DAT though.

 

[luder98]

Hey guys.. There are two probablity questions that I cannot figure out. Can someone PLEASE help!! Thanks guys. I am really starting to stress since I am retaking the DAT at the end of the month and I have not even started anything but math yet....

1. What is the probability of getting 6 tails out of nine tosses of a fair coin??

a) 21/128
b) 21/64
c) 84/128
d) 22/64
e) 42/64

The correct choice is A

Use formula prob = desired outcomes/total outcomes

Total outcomes: for each toss, there are two outcomes. Therefore, for 9 tosses, there are 2*2...*2 (9 times) = 2^9

Desired outcomes: This is a little bit tricky. Number of desired outcomes is equal to the combination of 9 and 6, which is

9!/[6!*(9-6)!] = 9!/(6!*3!) = 6!*7*8*9/(6!*3!) = 7*8*9/(2*3) = 7*4*3

Therefore, prob = 7*4*3/(2^9) = 21/(2^7) = 21/128

 

[frogger33]

Thanks guys! That helped! I really dont know where I would be without you guys! i will probably be back with more questions soon... 🙂