Probability that sum of four number equal 50.

Question

I have this question:

We choose 4 numbers randomly with replacement from the space $\{0,1,2,3,...,50\}.$

a) What is the probability that the sum of the numbers is exactly 50?

b) What is the probability that the sum of the numbers is exactly 50 and all numbers are bigger or equal to 10?

For (a): I tried to calculate the probability that the first three numbers are less or equal to 50, then multiplying the answer by $1/50.$

But I am stuck on calculating the probability. Is my approach good? any hints?

For (b): I don't have any clue what to do.

Answer 1

a)

The set $\Omega=\{0,1,2,\dots,50\}^4$ can be used as outcome space and has cardinality $51^4$.

The events $\{\omega\}$ where $\omega\in\Omega$ are equiprobable.

So the probability will equal $51^{-4}|A|$ where $A=\{\omega\in\Omega\mid \omega_1+\omega_2+\omega_3+\omega_4=50\}$.

The cardinality of $A$ can be found by means of stars and bars and equals $\binom{50+3}{3}$.

b)

Similar story, but now we are looking for the cardinality of $B=\{\omega\in\Omega\mid \omega_1+\omega_2+\omega_3+\omega_4=50\text{ and }\omega_i\geq10\text{ for } i=1,2,3,4\}$

Setting $\alpha_i=\omega_i-10$ for $i=1,2,3,4$ we find that $B$ has the same cardinality as the set $C=\{\alpha\in\Omega\mid\alpha_1+\alpha_2+\alpha_3+\alpha_4=10\}$

Again we can apply stars and bars. This time we find that $|B|=\binom{10+3}3$.

Answer 2

Denote your 4 numbers by: x$_1, x_2,x_3$ and $x_4$
Total number of selections is $$51^4$$

For a:
$$x_1+x_2+x_3+x_4=50$$ (where all $x_i$ lie in {0,1,,...,50})

So, using stars and bars method, number of solutions of above equation are: $$^{50+4-1}C_{4-1}=^{53}C_3$$ So, probability will be: $$\frac{^{53}C_3}{51^4}$$

For b: the set shrinks to {10,11,12,...,50}

So, $$x_1+x_2+x_3+x_4=50$$ (where all $x_i$ are $\geq10$)

So, replace all $x_i$ with $(10+y_i)$ such that $y_i \geq 0$ $\forall$ i $\in{1,2,3,4}$

So, $$(10+y_1)+(10+y_2)+(10+y_3)+(10+y_4)=50$$ Or, $$y_1+y_2+y_3+y_4=10$$ number of solutions is: $$^{10+4-1}C_{4-1}=^{13}C_3$$ So, probability will be: $$\frac{^{13}C_3}{51^4}$$