Good day, busy with the following problem. If you guys can please be so kind to direct me in the right direction. My solution doesn't seem right to me.
Question:
In a game of Lotto, balls are numbered 1 through to 44. They are all placed in a barrel and six balls are drawn without replacement. The balls are of the same size and are equally likely to be drawn. The first five balls drawn out are numbered 34, 2, 15, 29 and 42. What is the probability that the next ball drawn out will be number 26?
My first solution was:
After 5 balls were drawn that means only 39 balls left in the pot so the odds of picking 26 is 1 in 39 which results in 1/39 and a 2.5% chance.
But given that this is a statistics assignment I figured I most likely should do it using probability methods etc.
So my second solution was:
Event A: Ball 26 is drawn
Event B: Balls 34, 2, 15, 29 and 42 are drawn
P(A|B) = P(A and B) / P(B)
P(A|B) = (1/44) * (5/44) / (5/44)
P(A|B) = (5/1936) / (5/44)
P(A|B) = (5/1936) * (44/5)
P(A|B) = 220/9815
P(A|B) = 44/1963
P(A|B) = 0.022
2.2% chance of ball number 26 being drawn next.
Please assist 🙂
Best Answer
The probability of choosing $k$ specific cards from $44$ is $\frac1{\binom{44}{k}}$ since there are $\binom{44}{k}$ equally likely ways to choose $k$ cards. This gives a denominator of $1/\binom{44}5$. There are $\binom{44}6$ ways to choose $6$ specific cards, so the probability of drawing all $6$ is $$\frac1{\binom{44}{6}}$$ However, we are requiring that a specific card come last among the $6$, so the probability that we first draw $5$ specified cards and then a specific sixth card is $$\frac16\frac1{\binom{44}{6}}$$ Therefore, $$\Pr(A|B)=\frac16\frac{\binom{44}5}{\binom{44}6}=\frac1{39}$$
Another, perhaps clearer way to calculate $\Pr(A\cup B)$ is to use permuatations instead of combinations. There are $5!$ admissible draws, since we can choose the first five cards in any order, but the sixth is fixed. There are $$44\cdot43\cdot42\cdot41\cdot40\cdot39$$ so the proabability is $$\frac{5!}{44\cdot43\cdot42\cdot41\cdot40\cdot39}$$ which you can check is the same as before.