Math help? Probabilities?

[DoctaWalnuts]

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Does anyone have a great way of dealing with these probability questions?

Is there a source that you think would help someone that really is struggling with them?

 

[iwannabadentist]

can you perhaps post a practice question? I might be able to walk you thru it!

 

[nartnad]

i second that question. probabilities and anything that involves the use of factorials is my weak point. are there any good resources?


i have a specific question:

if you flip a coin 6 times, what is the probability that heads will show up exactly twice?

- the answer says 15/64. i can see that mabye you get that from 5!/2^6, but i dunno why you multiply by 5!

can somone explain?

 

[nartnad]

hmm nevermind what i said before. i found an answer....but i dont understand how a dat math question could be so complicated:


the chance of flipping a coin is .5, and since theres 6 flips, thats 1/64 chance of getting a desired outcome. i understand this. but to get the probability of flipping heads exactly TWICE, you multiply this by:

n!/[(n-r)!r!] where n = number of flips, and r = number of heads

so.... 6!/(6-2)!2! = 6!/4!2! = 15

so.... probability = 15/64

that seems very difficult. is there an easier way?

 

[nartnad]

another way: use pascuals triangle

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1

for 6 flips, use the row with the 6 in it:
1/64 probability of flipping 0 heads
6/64 " 1 head
15/64 " 2 heads
20/64 " 3 heads
etc.

i dont know which is faster/easier tho...

 

[EyEnStein 07]

hmm nevermind what i said before. i found an answer....but i dont understand how a dat math question could be so complicated:


the chance of flipping a coin is .5, and since theres 6 flips, thats 1/64 chance of getting a desired outcome. i understand this. but to get the probability of flipping heads exactly TWICE, you multiply this by:

n!/[(n-r)!r!] where n = number of flips, and r = number of heads

so.... 6!/(6-2)!2! = 6!/4!2! = 15

so.... probability = 15/64

that seems very difficult. is there an easier way?

The formula way you stated is probably the easiest i can think of. the only other way is to know that six flips in a quarter is 2^6 = 56 total amount of possibilities. From there it gets complicated because now you have to start
listing the combination...and 56 is quite a bit (for ex. if there were 4 flips..its much easier since theres only 16)

i would just advise to stick to that formula.