# probability - unsure of what type

• Apr 12th 2010, 03:19 PM
shawli
probability - unsure of what type
The probability that a certain dart player hits the bullseye with one dart is $0.4$.

a) Find the probability that the player scores at most two bullseyes with $3$ darts.
b) If the probability of scoring at least one bullseye with $n$ darts is greater than $0.9$, find the least possible value of $n$.
• Apr 12th 2010, 03:36 PM
pickslides
Quote:

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

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

Have you tried Binomial?

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

b) You will need to work backwards.
• Apr 12th 2010, 03:43 PM
shawli
It IS binomial, thank you!
• Apr 12th 2010, 04:52 PM
losm1
I tried to solve b):

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

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

$0.9 < 1 - 0.6^n$

$0.6^n < 0.1$

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

$n > 4.50$

I'm not sure about a), does $P(X \le 2) = 1 - P(one\ miss)$ and then $1 - {3 \choose 1} 0.4^2 0.6$ makes sense?
• Apr 12th 2010, 05:10 PM
pickslides
Quote:

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

My understanding is.

$P(X\leq 2) = P(X=0)+P(X=1)+P(X=2)=1-P(X=3)= 1 - {3 \choose 3} \times 0.4^3 \times 0.6^0$
• Apr 13th 2010, 12:34 AM
losm1
That is correct. Thanks