Answer to Question #138619 in Statistics and Probability for Cebb

Question

Construct the probability distribution if three cellphones were to be tested, consider the random variable (x) as the number of defected cellphones.

Answer 1

Let's denote probability that phone is defected as $p$. Then, it's the binomial distribution:

$Pr(x=k)= \binom{3}{k} p^{k} \cdot(1-p)^{3-k},$ where $k$ denotes number of defected phones.

Thus:

$Pr(x=0)= \binom{3}{0} p^{0} \cdot(1-p)^{3-0} = (1-p)^{3}$

$Pr(x=1)= \binom{3}{1} p^{1} \cdot(1-p)^{3-1} = 3 \cdot p \cdot (1-p)^{2}$

$Pr(x=2)= \binom{3}{2} p^{2} \cdot(1-p)^{3-2} = 3 \cdot p^{2} \cdot (1-p)$

$Pr(x=3)= \binom{3}{3} p^{3} \cdot(1-p)^{3-3} = 3 \cdot p^{3}$