22-ntb_bayesian_theorem.ipynb

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   "source": "\\begin{align*}\nP(A|B) &= \\frac{P(B \\cap A)}{P(B)}\\\\\n\\end{align*}\nwhere\n- $P(A)$ is the probability that event $A$ occurs.\n- $P(B)$ is the probability that event $B$ occurs.\n- $P(B \\cap A)$ is the probability that both event $A$ and $B$ occur.\n- $P(A|B)$ is the probability that event $A$ occurs given the event $B$ occurs.",
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   "source": "$D$: A patient has COVID.\n\n$H$: A patient doesn't have COVID.\n\n$+$: A patient has a positive test result.\n\n$-$: A patient has a negative test result.\n\nThe probability that a patient has a positive test result when he/she doesn't have COVID is\n\n$P(+|H) = 0.01.$\n\nThe probability that a patient has a positive test result when he/she has COVID is\n\n$P(+|D) = 0.98.$\n\nThe probability that a patient has COVID is\n\n$P(D) = 0.01$.",
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   "source": "For a randomly selected person, what is the probability that a patient has a positive test result?\n\n$P(+) =$ ?\n\\begin{align*}\nP(H) &= 1 - P(D)\\\\\n&= 1 - 0.01\\\\\n&= 0.99.\\\\\nP(+) &=P(+|D)P(D) + P(+|H)P(H)\\\\\n&= 0.98(0.01)+0.01(0.99)\\\\\n&= 0.0197.\n\\end{align*}",
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   "source": "If a randomly selected person is tested positive, what is the probability the person has COVID?\n\n$P(D|+)=$?\n\n\\begin{align*}\nP(D|+) &= \\frac{P(+\\cap D)}{P(+)}\\\\\n&= \\frac{P(+|D)P(D)}{P(+)}\\\\\n&= \\frac{0.98(0.01)}{0.0197}\\\\\n&= 0.497.\n\\end{align*}",
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   "source": "If a randomly selected person tested positive on each of two test applications, what is the probability that the person has the disease?",
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   "source": "We denote:\n- $+_1$: First test is positve\n- $+_2$: Second test is positive\n\nWe know $P(+_1|D) =98\\%$, $P(+_2|D) =98\\%$, $P(+_1|H) =1\\%$, $P(+_2|H) =1\\%$.",
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   "source": "Since $+_1$ and $+_2$ are independent, we have\n\\begin{align*}\nP(+_1 \\cap +_2|D) &= P(+_1|D) P(+_2|D)\\\\\n&= (0.98)(0.98)\\\\\n&= 0.9604.\n\\end{align*}\n\n\\begin{align*}\nP(+_1 \\cap +_2|H) &= P(+_1|H) P(+_2|H)\\\\\n&= (0.01)(0.01)\\\\\n&= 1\\times 10 ^{-4}.\n\\end{align*}\n\n\\begin{align*}\nP(+_1 \\cap +_2) &= P(+_1 \\cap +_2|D)P(D) +  P(+_1 \\cap +_2|H)P(H)\\\\\n&= 0.9604\\times 0.01 +  1\\times 10 ^{-4}\\times0.99\\\\\n&= 0.0097\n\\end{align*}\n\nWe have\n\n\\begin{align*}\nP(D|+_1 \\cap +_2) &= \\frac{P(D \\cap (+_1 \\cap +_2))}{P(+_1 \\cap +_2)}\\\\\n&= \\frac{P(+_1 \\cap +_2|D)P(D)}{P(+_1 \\cap +_2)}\\\\\n&= \\frac{0.9604 \\times 0.01}{0.0097}\\\\\n&= 0.9901.\n\\end{align*}",
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