Question

3.5. With reference to the following figure, find a) P(A B) b) P(BIC) c) P(A n B|C) d) P(B U CIA) e) P(AB u c) 0.06 0.24 0.19 0.04 0.16 0.11 0.11 0.09 3.6. For two rolls of a balanced die, find the probabilities of getting a) two 4s b) first a 4 and then a number less than 4

Answer 1

Solution :-

We will use Bayes rule to find the given probabilities...

For simplicity we will find following probabilities

$\\ {P(A)} = 0.24 + 0.06 + 0.04 + 0.16 = 0.5 \\ {P(\bar{A})} = 1 - {P(A)} = 1 - 0.5 = 0.5 \\ {P(B)} = 0.19 + 0.06 + 0.04 + 0.11 = 0.4 \\ {P(\bar{B})} = 1 - {P(B)} = 1 - 0.4 = 0.6 \\ {P(C)} = 0.09 + 0.11 + 0.04 + 0.16 = 0.4 \\ {P(\bar{C})} = 1 - {P(C)} = 1 - 0.4 = 0.6 \\$

${a)} P (A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.06 + 0.04}{0.19 + 0.06 + 0.04 + 0.11} = \frac{0.1}{0.4} = {0.25}$

${b)} P (B|\bar{C}) = \frac{P(B \cap \bar{C})}{P(\bar{C})} = \frac{0.06 + 0.19}{1 - (0.09 + 0.11 + 0.16 + 0.04)} = \frac{0.25}{0.6} \approx 0.42$

${c)} P (A \cap B|{C}) = \frac{P(A \cap B \cap C)}{P({C})} = \frac{0.04}{0.4} = 0.1$

${d)} P (B \cup C|\bar{A}) = \frac{P((B \cup C) \cap \bar{A})}{P(\bar{A})} = \frac{0.09 + 0.11 + 0.19}{0.5} = \frac{0.39}{0.5} = 0.78$

$\\ {e)} P (A | B \cup C) = \frac{P(A \cap (B \cup C) )}{P(B \cup C)} = \frac{0.16 + 0.04 + 0.06}{0.4 + 0.4 - 0.04 - 0.11} = \frac{0.26}{0.65} = 0.3846$

$\\ {f)} P (A | B \cap C) = \frac{P(A \cap (B \cap C) )}{P(B \cap C)} = \frac{0.04}{0.04 + 0.11} = \frac{0.04}{0.15} \approx 0.2667$

$\\ {g)} P (A \cap B \cap C | B \cap C) = \frac{P((A \cap B \cap C)\cap (B \cap C) )}{P(B \cap C)} = \frac{P(B \cap C)}{P(B \cap C)} = 1$

$\\ {h)} P (A \cap B \cap C | B \cup C) = \frac{P((A \cap B \cap C)\cap (B \cup C) )}{P(B \cup C)} = \frac{0.04}{0.65} \approx 0.06153$

Solution for 3.6 :-

In the case of rolling balanced die Probability of each digit is :- (1/6)

$\\ {a)} P ({two \ 4s}) = \frac{1}{6} * \frac{1}{6} = \frac{1}{36}$

$\\ {b)} P ({first \ 4 \ and \ num \ less \ than \ 4}) = \frac{1}{6} * \frac{3}{6} = \frac{3}{36} = \frac{1}{12}$