Probability ,balls from urn question

Quite confused, I want to know if im on the right track, and if so, how to explain it,

The question:

In an urn we have 7 black balls and 5 white balls, we draw a ball at **random ,and put it back in the urn time after time,what is the probability that on the 10th time we get a black ball is on the 20th draw?**

Intuitively im thinking that the probability is 7/12, because it doesnt matter if take a black ball first or 10th...but I really hope someone can direct me in the right direction.

• You are neglecting the "$10^{th}$ time" constraint. Yes, you need draw $20$ to be black but you also need exactly $9$ of the prior $19$ draws to be black. – lulu Oct 30 '18 at 16:39
• This is the same as asking what the probability is of getting the tenth head on the 20th toss of an unfair coin whose probability of getting a head is $\frac{7}{12}$. – John Douma Oct 30 '18 at 16:44
• To emphasize why the answer is not simply $\frac{7}{12}$, consider a related question about coins and asking for the probability that the third flip is a head and further that the third flip was the first time head happened. You have the eight equally likely events: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Yes, you have four of the eight times i.e. $\frac{1}{2}$ of the time the third flip is a head, but only one of those eight times is the third flip the first time head happens, namely the outcome TTH. – JMoravitz Oct 30 '18 at 17:03
• Thank you for your answers!,unfortunately Im still not quite getting it, I need to go back and re watch my material, but If someone could write an answer for this question though I would appreciate it – user3184910 Oct 30 '18 at 17:35

When answering such questions, you need to distinguish the event given, i.e. that has already occurred, and the event which followed, whose probability you have to find.

The event that has already occurred here would be - out of the first $$19$$ balls picked, $$9$$ were black. Lets call this event A.

Now event B would be that on the $$20^{th}$$ draw, we got a black ball. Since after every time you pick a ball, it is put back in the set, the two events are independent.

P(B/A) = P(A)P(B)

P(A)= $$\binom{19}{9}(\frac{7}{12})^{9}(\frac{5}{12})^{10}$$

P(B)=$$7/12$$

• Formatting tip: type $P(B \mid A)$ to obtain $P(B \mid A)$. – N. F. Taussig Oct 31 '18 at 12:24
• @N.F.Taussig noted...will do that next time. – Karan Abrol Oct 31 '18 at 18:18

From @lulu's comment, $$P(\,exactly\,\,9\,\,black\,balls\,in\,first\,\,19\,\,draws\, ) = \binom{19}{9}p^{9}(1-p)^{10}$$

$$P(\,black\, ball\,) = p = \frac{7}{12}$$

First, you must pick exactly 9 black balls out of the 19 draws, then you pick the last one on the 20th draw. Since these two events independent, we just multiply the two:

$$P(\,10th\,black\, ball\, on\, 20th\, draw\, ) = \binom{19}{9}p^{9}(1-p)^{10} * p \approx 0.0665$$

Which is what I get when I plug into wolframalpha. Sorry if my formatting looks weird, I am new to MSE and still learning how to format my answers. I hope this helps!

• If you type $P(\text{exactly}~9~\text{black balls in first}~19~\text{draws})$, you will obtain $P(\text{exactly}~9~\text{black balls in first}~19~\text{draws})$. – N. F. Taussig Oct 31 '18 at 12:25