Solution

 Let $T$ be the event that site has tear and  $M$ be the event that MRI is positive.

a. In table, we can see that $n(T \cap M)= 27$ and total observations, $n=70$

This gives $P(T \cap M)=\dfrac{n (T \cap M)}{n}= \dfrac{27}{70}\approx 0.38$

b. In table, we can see that $n(T^c \cap M)= 0$.

This gives $P(T^c \cap M)= \dfrac{n(T^c \cap M)}{n}=\dfrac{0}{70}=0.$

c. In table, we can see that $n(T \cap M^c)= 4$.

This gives $P(T \cap M)=\dfrac{n(T \cap M^c)}{n}= \dfrac{4}{70}\approx 0.06$

$M$

$T$

Answer

a. 0.38  b. 0 c. 0.06 d. 0.87 e. 0.13

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