Consider three sets E1={1,2,3},F1={1,3,4} and G1={2,3,4,5}. Two elements are chosen at random, without replacement, from the set E1and let S1denote the set of these chosen elements. Let E2=E1−S1 and F2=F1∪S1. Now two elements are chosen at random, without replacement, from the set F2 and let S2 denote the set of these chosen elements. Let G2=G1∪S2 . Finally, two elements are chosen at random, without replacement, from the set G2 and let S3 denote the set of these chosen elements.Let E3=E2∪S3 . Given that E1=E3 let "p" be the conditional probability of the event S1={2,3} . Then the value of "p" is

S1 F2=F1∪S1S2 G1∪S2=G2 S3 (i) {1,2} {1,2,3,4} {1,x} {1,2,3,4,5,} {1,2}{2,3}{1,2,3,4} {1,x} {1,2,3,4,5} ,{2,3} {x,y}( where x and y or {2,3,4,5,} (iii) {1,3}{1,3,4} {1,x} are other than 1) {1,2,3,4,5} {1,3} (i) P1= 2C2 3C2⋅ 3C1 4C2⋅ 2C1 5C2=160 (ii) P2=13 3C1 4C2× 2C2 5C2+ 3C2 4C2× 2C2 4C2=245 (iii) P3=1 3C1× 2C1 3C2× 2C2 5C2=145 Conditional probability =P2P1+P2+P3=815

Consider three sets E1={1,2,3},F1={1,3,4} and G1={2,3,4,5}. Two elements are chosen at random, without replacement, from the set E1and let S1denote the set of these chosen elements. Let E2=E1−S1 and F2=F1∪S1. Now two elements are chosen at random, without replacement, from the set F2 and let S2 denote the set of these chosen elements. Let G2=G1∪S2 . Finally, two elements are chosen at random, without replacement, from the set G2 and let S3 denote the set of these chosen elements.Let E3=E2∪S3 . Given that E1=E3 let "p" be the conditional probability of the event S1={2,3} . Then the value of "p" is

S1 F2=F1∪S1S2 G1∪S2=G2 S3 (i) {1,2} {1,2,3,4} {1,x} {1,2,3,4,5,} {1,2}{2,3}{1,2,3,4} {1,x} {1,2,3,4,5} ,{2,3} {x,y}( where x and y or {2,3,4,5,} (iii) {1,3}{1,3,4} {1,x} are other than 1) {1,2,3,4,5} {1,3} (i) P1= 2C2 3C2⋅ 3C1 4C2⋅ 2C1 5C2=160 (ii) P2=13 3C1 4C2× 2C2 5C2+ 3C2 4C2× 2C2 4C2=245 (iii) P3=1 3C1× 2C1 3C2× 2C2 5C2=145 Conditional probability =P2P1+P2+P3=815

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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 F₂and let S₂ 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 G₂ and let S₃denote 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 " $p$ " is

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$\frac{1}{5}$

$\frac{3}{5}$

$\frac{1}{2}$

$\frac{8}{15}$

answer is D.

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Detailed Solution

$S_{1} F_{2} = F_{1} \cup S_{1} S_{2} G_{1} \cup S_{2} = G_{2} S_{3}$
$(i) {1, 2} {1, 2, 3, 4} {1, x} {1, 2, 3, 4, 5,} {1, 2}$
${2, 3} {1, 2, 3, 4} {1, x} {1, 2, 3, 4, 5}, {2, 3} {x, y} (where x and y or {2, 3, 4, 5,} (iii) {1, 3} {1, 3, 4} {1, x} are other than 1) {1, 2, 3, 4, 5} {1, 3} (i) P_{1} = \frac{^{2} C_{2}}{^{3} C_{2}} \cdot \frac{^{3} C_{1}}{^{4} C_{2}} \cdot \frac{^{2} C_{1}}{^{5} C_{2}} = \frac{1}{60} (ii) P_{2} = \frac{1}{3} (\frac{^{3} C_{1}}{^{4} C_{2}} \times \frac{^{2} C_{2}}{^{5} C_{2}} + \frac{^{3} C_{2}}{^{4} C_{2}} \times \frac{^{2} C_{2}}{^{4} C_{2}}) = \frac{2}{45} (iii) P_{3} = \frac{1}{^{3} C_{1}} \times \frac{^{2} C_{1}}{^{3} C_{2}} \times \frac{^{2} C_{2}}{^{5} C_{2}} = \frac{1}{45} Conditional probability = \frac{P_{2}}{P_{1} + P_{2} + P_{3}} = \frac{8}{15}$