This post covers Introduction to probability from Statistics for Engineers and Scientists by William Navidi.

Exercises

• What is the probability that a randomly selected person is male, given that they own a pet?

Let $M$ is for male and $PO$ stands for pet owner, so the probability that a randomly selected male owns a pet is $P(M \mid PO) = P(M∩PO) / P(PO) = .41$

$P(PO) = .86$

$P(M \mid PO)= P(M \cap PO)/P(PO) = .41/.86 = .477$

• A juggler has seven red, five green, and four blue balls. During his stunt, he accidentally drops a ball and then picks it up. As he continues, another ball falls. What is the probability that the first ball that was dropped is blue, and the second ball is green?

  • The first ball is picked by the juggler therefore the size of the sample space for both balls is 16, because these events are independent.

    The probability that the first ball is blue or P (blue ball) = 4/16

    The probability that the second ball is green or P(green ball) = 4/16

    The probability that the first ball is blue and the second ball is green:

    P(blue and green)= P(blue) × P(green) = 4/16 × 4/16 = 1/16

    Thus, the probability is 1/16 for both cases.

• A man is known to speak the truth 2 out of 3 times. He throws a die and reports that the number obtained is a four. Find the probability that the number obtained is actually a four.

  • Let $A$ be the event that the man reports that number four is obtained. Let $E_{1}$ be the event that four is obtained and $E_2$ be its complementary event. Then,

    $P(E_1)$ = Probability that four occurs = $(\frac{1}{6})$ and

    $P(E_{2})$ = Probability that four does not occurs = $ (1 - P(E_{1}) ~=~ 1~-\frac{1}{6}~ = ~\frac{5}{6}$

    Also, $P(A \mid E_{1})$ = Probability that man reports four and it is actually a four = $(\frac{2}{3})$

    $P(A \mid E_{2})$ = Probability that man reports four and it is not a four = $(\frac{1}{3})$

    By using Bayes’ theorem, probability that number obtained is actually a four,

    $P(E_{1}|A) = \large \frac{P(E_1)P(A|E_1)}{P(E_1 )P(A│E_1 )~+~ P(E_2)P(A|E_2)}~ =~\frac{\frac{1}{6} ~ ×~ \frac{2}{3}}{\frac{1}{6} ~×~ \frac{2}{3}~ +~ \frac{5}{6}~ ×~\frac{1}{3}} = \frac{2}{7}$