# Math Help - Probability Question

1. ## Probability Question

Hello,

I am stuck on this probability question:

A fair coin is tossed 6 times. What is the probability of tossing exactly four heads in 6 tosses? What is the probability of tossing at most two heads. Each the first and second questions are independant of each other.

I believe I would know how to do it if if asked for a dice, because then the probabailty for each number is different. But the probability of heads or ails is both 50%. This is where I get stuck.

I would appreciate any help. Thank you.

2. Originally Posted by OnMyWayToBeAMathProffesor
Hello,

I am stuck on this probability question:

A fair coin is tossed 6 times. What is the probability of tossing exactly four heads in 6 tosses? What is the probability of tossing at most two heads. Each the first and second questions are independant of each other.

I believe I would know how to do it if if asked for a dice, because then the probabailty for each number is different. But the probability of heads or ails is both 50%. This is where I get stuck.

I would appreciate any help. Thank you.
this is a binomial distribution problem.

Suppose we have n independent trials, the outcome of which can be considered a success or failure, and we are looking for k successes each with a probability of p. we call the probability of failure q, where q = 1 - p, of course.

the probability of k successes is given by: $P(k) = {n \choose k}p^kq^{n - k}$

by the way, usually problems with dice are a bit more complicated than problems with coins

So I am grateful for your formula and I put it to use. This is what i got:

for the probability of tossing exactly 4 heads;

$P(4) = {6 \choose 4}(\frac{1}{2})^4(\frac{1}{2})^{6 - 4}$ and I received 23.4%. Is this correct?

for the probability of at most two heads;

$({6 \choose 1}(\frac{1}{2})^1(\frac{1}{2})^{6 - 1})+({6 \choose 2}(\frac{1}{2})^2(\frac{1}{2})^{6 - 2})$ and I received 32.8%. Is this correct?

Thank you.

4. Originally Posted by OnMyWayToBeAMathProffesor
So I am grateful for your formula and I put it to use. This is what i got:

for the probability of tossing exactly 4 heads;

$P(4) = {6 \choose 4}(\frac{1}{2})^4(\frac{1}{2})^{6 - 4}$ and I received 23.4%. Is this correct?
yes

for the probability of at most two heads;

$({6 \choose 1}(\frac{1}{2})^1(\frac{1}{2})^{6 - 1})+({6 \choose 2}(\frac{1}{2})^2(\frac{1}{2})^{6 - 2})$ and I received 32.8%. Is this correct?

Thank you.
no. of course you realize "at most two heads" means two or less. what you forgot is that zero heads is a possible outcome that is less than two heads