Respuesta :
Answer: The required probabilities is
(a) 0.504 (b) 0.054 (c) 0.092.
Step-by-step explanation: Given that Kevin will soon be taking exams in math, physics, and French. He estimates the probabilities of his passing these exams to be as follows:
[tex]P(m)=0.9,~~~P(p)=0.8,~~~P(f)=0.7.[/tex]
The results of these three exams are independent of each other.
We know that if A and B are independent events, then
[tex]P(A\cap B)=P(A)\times P(B).[/tex]
(a) The probability that Kevin pass all three exams is
[tex]P(m\cap p\cap f)=P(m)\times P(p)\times P(f)=0.9\times0.8\times0.7=0.504.[/tex]
(b) The probability that Kevin pass math but fail the other exams is
[tex]P(m\cap \bar{p}\cap \bar{f})=P(m)P(\bar{p})P(\bar{f})=0.9(1-0.8)(1-0.7)=0.9\times0.2\times0.3=0.054.[/tex]
(c) The probability that Kevin pass exactly one of the three exams is
[tex]P(m\cap \bar{p}\cap \bar{f})+P(\bar{m}\cap p\cap \bar{f})+P(\bar{m}\cap \bar{p}\cap f)\\\\=0.9(1-0.8)(1-0.7)+(1-0.9)0.8(1-0.7)+(1-0.9)(1-0.8)0.7\\\\=0.054+0.024+0.014\\\\=0.092.[/tex]
Thus, the required probabilities is
(a) 0.504 (b) 0.054 (c) 0.092.
