# Thread: probability - unsure of what type

1. ## probability - unsure of what type

The probability that a certain dart player hits the bullseye with one dart is $\displaystyle 0.4$.

a) Find the probability that the player scores at most two bullseyes with $\displaystyle 3$ darts.
b) If the probability of scoring at least one bullseye with $\displaystyle n$ darts is greater than $\displaystyle 0.9$, find the least possible value of $\displaystyle n$.

2. Originally Posted by shawli
The probability that a certain dart player hits the bullseye with one dart is $\displaystyle 0.4$.

a) Find the probability that the player scores at most two bullseyes with $\displaystyle 3$ darts.
b) If the probability of scoring at least one bullseye with $\displaystyle n$ darts is greater than $\displaystyle 0.9$, find the least possible value of $\displaystyle n$.
Have you tried Binomial?

a) find $\displaystyle P(X\leq 2)$ where $\displaystyle X\text{~}Bi(n,p) = X\text{~}Bi(3,0.4)$

b) You will need to work backwards.

3. It IS binomial, thank you!

4. I tried to solve b):

$\displaystyle P(X \ge 1) = 1 - P(X = 0)$

$\displaystyle 0.9 < 1 - P(X = 0)$

$\displaystyle 0.9 < 1 - 0.6^n$

$\displaystyle 0.6^n < 0.1$

$\displaystyle n > \frac{ln(0.1)}{ln(0.6)}$

$\displaystyle n > 4.50$

I'm not sure about a), does $\displaystyle P(X \le 2) = 1 - P(one\ miss)$ and then $\displaystyle 1 - {3 \choose 1} 0.4^2 0.6$ makes sense?

5. Originally Posted by losm1
I'm not sure about a), does $\displaystyle P(X \le 2) = 1 - P(one\ miss)$ and then $\displaystyle 1 - {3 \choose 1} 0.4^2 0.6$ makes sense?
My understanding is.

$\displaystyle P(X\leq 2) = P(X=0)+P(X=1)+P(X=2)=1-P(X=3)= 1 - {3 \choose 3} \times 0.4^3 \times 0.6^0$

6. That is correct. Thanks