1

I'm having a little trouble with a homework problem. I will write out what I've figured out and would appreciate some help with what I'm unable to understand.

You flip a weighted coin four times. Assume that Pr [Heads]=0.7 and Pr [Tails]=0.3. The random variable X is defined to be 4 times the number of heads minus 1 times the number of tails.

So landing four heads would result in an X value of 16 and the probability is .7^4. And landing four tails would result in an X value of -4 and the probability is .3^4. I have the X values for the other three scenarios but I'm having trouble calculating their probabilities. I know that the probability values for all events must add up to 1.

So for example, if I flip 1 heads and 3 tails, X=1. I was trying .7*.3^3 to find the probability, but I realize I'm making a mistake in this calculation. Please help. Thank you.

Tyranitator
  • 57

1 Answers1

1

Let $Y$ be the number of heads. Then the number of tails is $4-Y$, and $X=4Y-(4-Y)=5Y-4$.

The values taken on by $X$, that is, by $5Y-4$, are $-4$, $1$, $6$, $11$, and $16$. We find $\Pr(X=a)$ for $a=-4$, $1$, $6$, $11$, and $16$.

$\Pr(X=-4)$: This is the probability of $0$ heads in $4$ tosses. This probability is $(0.3)^4$.

$\Pr(X=1)$: This is the probability of $1$ head in $4$ tosses. The binomial distribution tells us that the probability of $1$ head and therefore $3$ tails is $\binom{4}{1}(0.7)^1(0.3)^3$.

$\Pr(X=6)$: This is the probability of $2$ heads and therefore $2$ tails. It is $\binom{4}{2}(0.7)^2(0.3)^2$.

Similarly, $\Pr(X=11)=\binom{4}{3}(0.3)^3(0.7)^1$ and $\Pr(X=16)=(0.7)^4$.

Remark: Suppose we repeat an experiment independently $n$ times, and the probability of success on any trial is $p$, and the probability of failure is $1-p$. Then the number $W$ of successes in the $n$ trials has binomial distribution, and $\Pr(W=w)=\binom{n}{w}p^w(1-p)^{n-w}$.

André Nicolas
  • 507,029