Question

There are two therapies $B_1$ and $B_2$ available for curing a patient suffering from a certain disease. The patient can choose any one of the two therapies. If he selects therapy $B_1$ the probability of his recovery from the disease is $\frac{7}{8}$ and if he selects therapy $B_1$ the probability of his recovery from the disease is $\frac{9}{10}$. What will be the probability if he chooses any one of the therapies.

Answer 1

Given that:

$P\left(B_1\right)=\frac{7}{8}$

$P\left(B_2\right)=\frac{9}{10}$

$P(B_1\bigcap B_2)=P\left(B_1\right)\times P\left(B_2\right)$

$P(B_1\bigcap B_2)=\frac{7}{8}\times \frac{9}{10}$

$P(B_1\bigcap B_2)=\frac{63}{80}$

$P(B_1\bigcup B_2)=P\left(B_1\right)+P\left(B_2\right)$$-P(B_1\bigcap B_2)$

$P(B_1\bigcup B_2)=\frac{7}{8}+\frac{9}{10}-\frac{63}{80}$

$P(B_1\bigcup B_2)=0.875+0.900-0.7875$

$P(B_1\bigcup B_2)=0.9875$

Hence, the required probability is

$P(B_1\bigcup B_2{)}^{\prime }=1-0.9875=\mathrm{0.0125.}$