Nine fair coins are tossed at once.?

Question

a. What is the probability that exactly one is a head?

b. What is the probability that exactly two are heads?

c. What is the probability that at most two are heads?

d. What is the probability that at least two are heads?

Answer 1

If you have been shown how to create Pascal's triangle, this one is a slam dunk. Carry it down to the ninth row (the one that starts 1 9 36 84 ...) and then you can answer all those questions with at worst a little addition.

Answer 2

A) There are 9 coins, with two sides (obviously) so the probability of any coin being heads or tails is 1/2. If you had two coins, the probability of there being both heads/tails would be 1/2*1/2+1/4 and the probability of there being one heads and one tails would be 2*1/2*1/2+1/2, since you can have T-H or H-T. Thus, the chance of, with nine coins, having only a single head show up would be:
1/2*1/2*1/2...nine times...*1/2*9 (nine different chances of getting H)= 9/512

B) With 512 possible different (not unique) outcomes, then chance of getting exactly 2 H would be 1/512*36 (combinations of getting two H) = 36/512.

C) This could be rephrases to say: What is the probability that either one or two heads are shown? This is a combiniation of questions A and B, with the inclusion of no heads (only one outcome). The chance of getting one H is 9/512 while the chance of getting 2 H are 36/512. Thus, the probability that either can happen is 46/512.

D) This questions is accounting for every combination with the exception of if a single (or no) head is shown. This is found by including the possibility of getting all tails to answer A. Thus there are only 10 outcomes that do now have at least two heads. Thus the answer is 502/512.

Answer 3

For each situation, the same base formula applies, P = m+n Choose n * p1^n * p2^m

For example, For 9 coins with 1 head and 8 tails,

P = 9 choose 1 * 1/2^1 * 1/2^ 8.

or 9/512

For 2 heads, it is 9 choose 2 * 1/2^9
since you said they are fair coins you will always have 2^9 in the denominator, so,
36/512

In order to answer part c, we need the prob of no heads = 9 choose 0 * 1/2^9 or 1/512

So, in order to have "at most" two heads, you can have 0, 1, or 2 and the prob of one of these is 1/512 + 9/512 + 36/512 = 46/512

Finally, the prob of getting "at least 2 heads" is 1 minus prob of getting 0 or 1 head which is
1 - 1/512 - 9/512 = 502/512

Answer 4

I thought that it is always 50/50 since there ar only two sides to the coins. But then I am just a non-mathematical person, and your qu is proabaly about some rules and regulations that govern such things. thanks for 2 points

Answer 5

a.] 1/18
b.] 1/9
c.] huh
d.] yeah.... never mind i thought i knew but i dont mybad