Question 1182155
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There are 2 possible outcomes on each toss of the coin.  The total number of possible outcomes is 2*2*2*2*2=32.  It looks as if you did 2+2+2+2+2=10.<br>
Make sure you understand why you multiply the 2's instead of adding them.  It's the fundamental counting principle: if there are m ways to do one thing and n ways to do another thing (and those two things are independent), then the number of ways to do those two things one after the other (or at the same time) is m times n.<br>
In this problem, you are doing 5 things (the 5 coin tosses), each of which has 2 possible outcomes, so you multiply all those 2's together to get the total number of possible outcomes.<br>
There are not only 3 ways to get 3 heads in 5 tosses.  The number of ways of getting 3 heads in 5 tosses is "5 choose 3", often denoted by either C(5,3) or 5C3 (we need to choose 3 of the 5 tosses to be heads).  Mathematically, the computation for 5C3 is<br>
{{{5!/((3!)(5-3)!)=5!/((3!)(2!)) = (5*4*3*2*1)/((3*2*1)(2*1)) = (5*4)/(2*1)=10}}}<br>
So the probability for this first problem is 10/32 = 5/16.<br>
The other problems are worked in exactly the same way:
n tosses means 2^n possible outcomes
m heads (or tails) in n tosses can happen in nCm ways
The probability of m heads (or tails) in n tosses is {{{(nCm)/(2^n)}}}<br>
A couple of things to keep in mind when you are calculating nCm values:<br>
(1) As an example, 10C7 is the same as 10C3.<br>
10C3 means you are choosing 3 of the 10 items; that means you are choosing NOT to choose 7 of the 10 items.  So 10C7 is the same number as 10C3.<br>
(2) After you have performed the nCm calculation a few times for different problems, stop going clear back to the definition of nCm. Here is an example:<br>
To calculate 10C7, first change that to 10C3, since those are the same number.
Then the computation for 10C3 is a fraction with 3 factors in both numerator and denominator; the numbers in the numerator start with 10 and count down, and the numbers in the denominator start with 3 and count down:
{{{10C7=10C3=((10*9*8)/(3*2*1))=720/6=120}}}<br>
I'll show the calculation for one more of your problems as a further example.<br>
(d) 2 heads in 4 tosses<br>
The number of possible outcomes is 2^4=16
The number of ways of getting 2 heads in 4 tosses is {{{C(4,2) = (4*3)/(2*1)=12/2=6}}}
The probability is 6/16 = 3/8<br>
I leave (b) to you; you know what the answer is supposed to be, so you can check your work.<br>
Note that the answer for (c) is going to be the same as the answer to (b) -- because in (b) the numerator of the probability fraction is 8C2 and in (c) the numerator is 8C6.  But (from the first note above) 8C6 and 8C2 are the same number.<br>
For (e) the probability is going to be {{{C(20,10)/2^20}}}.  Those are not calculations you want to do by hand; and you can see the answer is "ugly".  Find a calculator to verify the given answer for that problem.<br>
Or, if you now feel comfortable with the process for finding the answers for problems like this, simply skip the last example.  (Working with the big ugly numbers won't teach you anything more about the PROCESS for finding the answer).<br>