Probability for a specific value for sum of normally distributed random variables

Question

I am trying to solve an exercise using the probability density function for a sum of n random variables with the same mean and variance. I need to find the probability for the specific value of the sums of r.v.s:

Player B uses a fair coin to earn points, if it lands on heads he collects 1 point, if tails he collects 8 points. When he flips the coin 50 times, these points will add up cumulatively.

What is the probability that player B will have exactly 225 points?

I was thinking I could take the $PDF$ probability for $\sum^n_{i=1}(X_i)=225$ with a mean of $n*µ=225$ and standard deviation of $\sqrt{\sigma^2*n}=24.75$, which gives the value 0.016, as can be seen on the PDF plot:

However, my teacher solved this by looking at it logically, and counting any situation where Player B gets exactly 25 heads and 25 tails:

$P=($$50\atop25$$)*\frac{1}{2^{50}}=0.112$

I see that his answer makes sense and is correct, but I doubt I would be able to think logically in an exam situation.

Why is my answer not correct, is it because the r.v. is only approximately normally distributed?

Is there another general way of solving such an exercise if the amount of flips and the total sum of points had been different?

Answer 1

Your method works fine, but you have to remember the continuity correction. You are approximating a discrete distribution with a continuous one. Thus, for instance, your method gives a non-zero chance of getting $224$ points (say) though that is in fact impossible.

As you are working with the scores, then for $(24,25,26)$ tails you get $(218,225,232)$ as scores...so for your continuous variant you need to look between $225-3.5$ and $225+3.5$. If you do that you get $0.112462916$ as opposed to the exact value of $0.112275173$ ... not bad at all!

Answer 2

The values $f(x)$ where $f$ denotes a PDF cannot be interpreted as probabilities (as you seem to think). Note for instance that they can take values that exceed $1$.

That is the mistake you made.

We have the equalities $H+T=50$ and $H+8T=225$ here leading to $H=25=T$, so the question can be rephrased as:

"By $50$ throws of a fair coin what is the probability on $25$ tails?"

This is the route taken by your teacher.

Answer 3

The first thing to notice is that $B$ throws 50 coins and makes 225 points. If $n_H$ and $n_T$ are the numbers of heads and tails, 50 throws make it so that $$ n_H + n_T = 50.$$ Given that a heads gives 1 point and a tails 8 points, we can set up the equation $$ 1 \cdot n_H + 8 \cdot n_T = 225.$$

Solving the two equations together gives $n_t = 25$; so we are after the probability tha 25 tails pop up in 50 throws. You can either do it with combinations and permutations (as your teacher did), or plainly use the pmf for the binomial distribution that represents your situation solving for 25 successes in 50 trials.

If you want the graphics, below is the plot of the distribution $B(50,0.5)$ with a line through $0.11227...$ which is the required probability.

While it is true that the binomial distribution tends to a normal distribution, you cannot interpret the values of a probability density function as the probability of an event.