Minimum times ball must be drawn A bag contains 2 red, 3 white and 5 black balls, a ball is drawn its
colour is noted and replaced. Minimum number of times, a ball must
be drawn so that the probability of getting a red ball for the first time
is atleast even.
I am not getting any start can anybody provide me a hint
 A: So the probability of drawing red is $1/5$ because 2 of the 10 balls are red.
You do this once, and the probability is $1/5$.
The probability of drawing at least one red in two tries is:
$$
{(1/5)}^2 + 2*{1/5}*{4/5} = 1/25 + 8/25 = 9/25
$$
(the chance you get red twice and the change you get red-other or other-red)
Still $<1/2$ so we're not there yet
for 3 draws, the same probability is
$$
{1/5}^3 + 3*{(1/5)}^2*4/5 + 3*{1/5}*{(4/5)}^2 = 61/125 = 0.488
$$
Still a bit short, so we go for the fourth try and get
$$
1/5^4 + 4*{(1/5)}^3*4/5 + 6*{(1/5)}^2*{(4/5)}^2 + 4*{1/5}*{(4/5)}^3 = 369/625 = 0.5904
$$
So we have that the # of times you have to do it is 4.
However, it is easier to go by "what's the probability that I don't get a red ball on $n$ many draws?" That probability is simply ${4/5}^n$ and then the probability that you get at least one red is $1-{(4/5)}^n$. Set that $>1/2$ and solve for $n$
$$
1-{(4/5)}^n > 1/2
$$
$$
{(4/5)}^n < 1/2
$$
$$
n\ln{(4/5)} < \ln (1/2)
$$$$n > \ln(1/2)/\ln(4/5) \approx 3.1062 \dots
$$
So $n=4$
