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rsansiya111
asked
in Probability
Dec 18, 2021

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1 vote

The required value is

P($H_{G}$|$H_{D}$) = $\frac{P(H_{G} ∩ H_{D})}{P(H_{D})}$

Now given $P(H_{D} | H_{G})$ = 0.4 = $\frac{P(H_{D} ∩ H_{G})}{P(H_{G})}$

$P(H_{D} ∩ H_{G})$ = $P(H_{D} | H_{G}) * P(H_{G})$

= 0.4 * 0.2 = 0.08

Now $P(H_{D})$ = $P(H_{D} | H_{G}) * P(H_{G}) + P(H_{D} | M_{G}) * P(M_{G})$

+ $P(H_{D} | L_{G}) * P(L_{G})$

= 0.4 * 0.2 + 0.1 * 0.5 + 0.01 * 0.3

= 0.133

Hence

P($H_{G}$|$H_{D}$) = $\frac{0.08}{0.133}$ = 0.6 (Corrected to 2 decimal places)

P($H_{G}$|$H_{D}$) = $\frac{P(H_{G} ∩ H_{D})}{P(H_{D})}$

Now given $P(H_{D} | H_{G})$ = 0.4 = $\frac{P(H_{D} ∩ H_{G})}{P(H_{G})}$

$P(H_{D} ∩ H_{G})$ = $P(H_{D} | H_{G}) * P(H_{G})$

= 0.4 * 0.2 = 0.08

Now $P(H_{D})$ = $P(H_{D} | H_{G}) * P(H_{G}) + P(H_{D} | M_{G}) * P(M_{G})$

+ $P(H_{D} | L_{G}) * P(L_{G})$

= 0.4 * 0.2 + 0.1 * 0.5 + 0.01 * 0.3

= 0.133

Hence

P($H_{G}$|$H_{D}$) = $\frac{0.08}{0.133}$ = 0.6 (Corrected to 2 decimal places)