Say you grab a card out a deck, check it, and put it back.
If the probability of getting a non-club card is $\frac{3}{4}$, what is the probability it will happen at least once out of two tries?
You can't just multiply it by $2$ because it can't be $>1$, so what do you do?
$\endgroup$ 12 Answers
$\begingroup$There are 3 possible scenarios:
- It happens at the first try (probability of $\frac{3}{4}$). Then it doesn't matter if you try again
- It doesn't happen the first try, but on the second try.
- It doesn't happen the first try, and also not on the second try.
The probability that either 2. or 3. occurs is $1-\frac{3}{4} = \frac{1}{4}$, i.e. that you got a club card on the first try.
Given that this is the case, the probability of getting non-club again is $\frac{3}{4}$.
So the overall probability of getting non-club either on the first try, or non on the first try but on the second try, is: $\frac{3}{4} + \frac{1}{4} \times \frac{3}{4} = \frac{15}{16} \approx 0.9375$
$\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$ would be the probability you get a non-club card twice. (And not just at least once, so stronger criterion with lower probability).
$\endgroup$ 1 $\begingroup$Method 1:
Ways in which you can draw at least one club
$\mathbb{P}($draw non club on the first card, then non club on second $)=\frac{3}{4}\cdot\frac{3}{4}=\frac{9}{16}$
$\mathbb{P}($draw non club on the first card, then club on second$)=\frac{3}{4}\cdot\frac{1}{4}=\frac{3}{16}$
$\mathbb{P}($draw club on first card, then non club on second$)=\frac{1}{4}\cdot\frac{3}{4}=\frac{3}{16}$
So total probability is:
$\mathbb{P}($probability at least one club drawn$)=\frac{9}{16}+\frac{3}{16}+\frac{3}{16}=\frac{15}{16}$
Method 2:
One can also notice that this probability is $1-\mathbb{P}($club drawn both times$)$
$\mathbb{P}($club drawn both times$)=\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16}$
So, $\mathbb{P}($probability at least one club drawn$)=1-\frac{1}{16}=\frac{15}{16}$
$\endgroup$ 1
