Consider three sets $E_{1} = {1, 2, 3}, F_{1} = {1, 3, 4}$ and $G_{1} = {2, 3, 4, 5}$ . Two elements are chosen at random, without replacement, from the set $E_{1}$ and let $S_{1}$ denote the set of these chosen elements. Let $E_{2} = E_{1} - S_{1}$ and $F_{2} = F_{1} \cup S_{1}$ . Now two elements are chosen at random, without replacement, from the set $F_{2}$ and let $S_{2}$ denote the set of these chosen elements. Let $G_{2} = G_{1} \cup S_{2}$ . Finally, two elements are chosen at random, without replacement, from the set $G_{2}$ and let $S_{3}$ denote the set of these chosen elements. Let $E_{3} = E_{2} \cup S_{3}$ . Given that $E_{1} = E_{3}$ , let p be the conditional probability of the event $S_{1} = {2, 3}$ . Then the value of 15 p is

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Detailed Solution

$S_{1}$ $F_{2} = F_{1} \cup S_{1}$ $S_{2}$ $G_{1} \cup S_{2} = G_{2}$ $S_{3}$

{1, 2} {1, 2, 3, 4} {1, x} {1, 2, 3, 4, 5,} {1, 2}
(ii) {2, 3} {1, 2, 3, 4} {1, x} {1, 2, 3, 4, 5,} {2, 3}
{x, y}(where x and or {2, 3, 4, 5,}
y are other than 1)
(iii) {1, 3} {1, 3, 4} {1, x} {1, 2, 3, 4, 5} {1, 3}
(i) $P_{1} = \frac{^{2} C_{2}}{^{3} C_{2}} . \frac{^{3} C_{1}}{^{4} C_{2}} . \frac{^{2} C_{1}}{^{5} C_{2}} = \frac{1}{60}$

(ii) $P_{2} = \frac{1}{3} (\frac{^{3} C_{1}}{^{4} C_{2}} \times \frac{^{2} C_{2}}{^{5} C_{2}} + \frac{^{3} C_{2}}{^{4} C_{2}} \times \frac{^{2} C_{2}}{^{4} C_{2}}) = \frac{2}{45}$