Let E1,E2,E3 and E4 be four events such that PEiE4=23,i=1,2,3 ; PEi∩EjcE4=16,i,j=1,2,3, i≠j,and P(E1∩E2∩E3c∣E4)=16 Then the value of PE1∪E2∪E3E4 is equal to

a+b+c+d=23 a+d=16b+e+f+c=23 c+f=16 c=16a+b+e+g=23 c+d=16d=0;a=1/6;e=1/6f=0;b=1/3;g=0a+b+c+d+e+f+g=23+16=56

Let E1,E2,E3 and E4 be four events such that PEiE4=23,i=1,2,3 ; PEi∩EjcE4=16,i,j=1,2,3, i≠j,and P(E1∩E2∩E3c∣E4)=16 Then the value of PE1∪E2∪E3E4 is equal to

a+b+c+d=23 a+d=16b+e+f+c=23 c+f=16 c=16a+b+e+g=23 c+d=16d=0;a=1/6;e=1/6f=0;b=1/3;g=0a+b+c+d+e+f+g=23+16=56

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Let $E_{1}, E_{2}, E_{3} and E_{4}$ be four events such that
$\begin{array}{rcl} P (\frac{E_{i}}{E_{4}}) & = & \frac{2}{3}, i = 1, 2, 3; \\ P (\frac{E_{i} \cap E_{j}^{c}}{E_{4}}) & = & \frac{1}{6}, i, j = 1, 2, 3, i \neq j \end{array}$
, $and P (E_{1} \cap E_{2} \cap E_{3}^{c} ∣ E_{4}) = \frac{1}{6}$

Then the value of $P (\frac{E_{1} \cup E_{2} \cup E_{3}}{E_{4}})$ is equal to

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$\frac{5}{6}$

$\frac{7}{12}$

$\frac{2}{3}$

$\frac{1}{2}$

answer is C.

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$a + b + c + d = \frac{2}{3} a + d = \frac{1}{6}$

$b + e + f + c = \frac{2}{3} c + f = \frac{1}{6} c = \frac{1}{6}$

$a + b + e + g = \frac{2}{3} c + d = \frac{1}{6}$

$d = 0; a = 1 / 6; e = 1 / 6$

$f = 0; b = 1 / 3; g = 0$

$a + b + c + d + e + f + g = \frac{2}{3} + \frac{1}{6} = \frac{5}{6}$

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Let E1,E2,E3 and E4 be four events such that PEiE4=23,i=1,2,3 ; PEi∩EjcE4=16,i,j=1,2,3, i≠j,and P(E1∩E2∩E3c∣E4)=16 Then the value of PE1∪E2∪E3E4 is equal to

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