Statistic problem concerning a coin

[UFStudent]

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Hello,

What is the best way solving coin toss problems?

1. Odds landing 3 heads out of 4 tosses?

2. Odds landing heads 6/10 tosses?

Thanks.

 

[DrTacoElf]

Bump for a good question.

 

[busdriver]

it's a good question...do you know the answer already? i have some ideas about the answer but i'm not 100% sure...

either way, i honestly don't think you'll get something like this on the DAT...just my opinion...(i'm not saying you won't get probability...i'm just saying that it won't be of this nature...)

 

[Mo007]

Hello,

What is the best way solving coin toss problems?

1. Odds landing 3 heads out of 4 tosses?

2. Odds landing heads 6/10 tosses?

Thanks.

This is a probability problem. Refer to this thread to get an idea how to solve the problems: http://forums.studentdoctor.net/showthread.php?t=117122

On top of my head, I think the answers are (not exactly sure):

1. 25% chance.

2. 21% chance.

What were the choices given to each problem?

 

[HBomb]

Hello,

What is the best way solving coin toss problems?

1. Odds landing 3 heads out of 4 tosses?

2. Odds landing heads 6/10 tosses?

Thanks.

Let me take a shot. Please anyone let me know if this is incorrect:
Use the binomial formula (I think that's what it's called) where C(x,y) is a combination, p is for probability of coin flip where p is 0.5 for a fair coin.
Probability = C(x,y) * p^y * (1-p)^(x-y)
FYI, the definition for C(x,y) is x!/[y!*(x-y)!]

For problem 1, if it's for exactly 3 heads and the coin is fair:
C(4,3) * (0.5)^3 * (0.5)^1 = 0.25

If you're more visual, here is the problem in binary (x denotes 3 heads):
0000
0001
0010
0011
0100
0101
0110
0111x
1000
1001
1010
1011x
1100
1101x
1110x
1111

For problem 2, do the same using 10 and 6.
You get C(10,6)*(0.5)^10 = 210/1024 = 0.205

Good luck.

 

[Mr Reddly]

A similar way to the visual (basically the same, just a slight tweak) approach above for number 1 would be not to think about 3 heads, but the fact that you have to get exactly 1 tails. This makes the problem alot easier to visualize. Lets do just that:

Thhh . 1st flip
hThh . 2nd flip
hhTh . 3rd flip
hhhT . 4th flip

As you can see (and it could have been done in your head), their are only 4 possible ways to get exactly 1 tails.
Great, we have the numerator.... 4
What about the denominator? That's just the total number of possible combinations,
2 possible outcomes for 4 flips is 2^4 = 16

So, the answer = 4 / 16 = 1/4 = 25%

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As for #2... Ya, what he said. 😀
Other than that, you know it's got to be smaller than 25% due to #1.
Thus, on a multiple choice test, you might be able to eliminate. 😳

 

[HBomb]

As Reddly says here, process of elimination might be a great way to do this problem depending on what the answer choices look like. If answer choices are improper fractions, then it might be more difficult, but generally, if you can use POE, it is a very sound idea to do so.

If you're kind of weak in QR and you have trouble finishing this section, it _might_ be a good idea to take your best quick guess, mark, and move on (and come back to it later if you have time). There are only a handful of stats questions (probably something like 2 or 3), so if you already have trouble finishing and your stats skills aren't good, why not just mark it and come back to it later?

But if you understand combinations and that binomial formula, you are golden. You can apply it to any of these coin flip problems. Try it a couple of times using the formula, and it will make sense. It's actually very logical, and you'll see the pattern for why it is the way it is. Plus, if the question is on the DAT, the coin is a fair coin (at least I have not seen a situation where it is not a fair coin)...so it makes that binomial equation even simpler.