Suppose you are playing darts where you get three throws. Let X be the number of times you hit a bullseye and have the following distribution table: hitting a dartboard .
What would x=2's probability be?
This is a binomial distribution question.
What would x=2's probability be?
0
$\begingroup$
probability
-
2What are your attempts at solving this? – 2017-02-10
-
0Probabilities must add to 1. – 2017-02-10
2 Answers
1
It is easier than it looks.
Note that there are only $3$ throws, therefore a maximum of $3$ possible bullseyes. All possible outcomes' probabilities must add up to $1$. Therefore, we have: $$P(X=0)+P(X=1)+P(X=2)+P(X=3)=1$$ Plug in the values you have for each probability, and solve for $P(X=2)$.
0
For Probability that it hits 2 times i.e when X=2 is
P(X=2) = 3 C (2) (probability that it hits right)^2 * (1-Probability that it hits right)^1