Most easy way is -
1) Total number of ways to select 2 boys and 2 girls from 6 - C(3,2) * C(3,2) - Desired outcomes = 3*3
2) Possible outcomes - C(6,4) = C(6,2) = 15
Probability = 3*3/15 = 3/5

Way you solved - considering probability of getting 2 boys * probability of getting 2 girls.
Mistake you did - Did not take possible outcomes of selecting boy from group of 6. Same for girl, and directly considering 2 outcomes.
Correct ways is - Conditional probability.
Probability of selecting 2 boys and 2 girls = P(1 boy) * P(2nd boy) * P(1 girl) * P(2nd girl).......(1) {BBGG}
... this can occur in many ways - BGBG, GGBB.... so you should multiply the number of ways you can get this combination of boy girl with the conditional probability (i.e. (1))

Below probability is conditional -
P(1 boy) * P(2nd boy) * P(1 girl) * P(2nd girl) = (3/6)*(2/5)*(3/4)*(2/3) = 1/10
Number of ways such pattern i.e. BGBG can occur is selecting 2 out of 4= C(4,2) = 4!/2!*2! = 6
Required probability = 6*1/10 = 3/5

beat_gmat_09 wrote: Number of ways such pattern i.e. BGBG can occur is selecting 2 out of 4= C(4,2) = 4!/2!*2! = 6
Required probability = 6*1/10 = 3/5

Thanks for the explanation.
The 1st method is clear.
But didn't understand the above part of 2nd method
can u plz explain a bit?

prachich1987 wrote:

beat_gmat_09 wrote: Number of ways such pattern i.e. BGBG can occur is selecting 2 out of 4= C(4,2) = 4!/2!*2! = 6
Required probability = 6*1/10 = 3/5

Thanks for the explanation.
The 1st method is clear.
But didn't understand the above part of 2nd method
can u plz explain a bit?

think of it as - in how many ways can you arrange 2 boys and 2 girls
which is - c(4,2) =6
or you can try to list the possibilities as-
bgbg
bbgg
gbgb
ggbb
bggb
gbbg
the value of conditional probability will be same in any particular combination, actually the probabilities of the events are added, but as the value of fraction remains same you can multiply the number of times you have to add I.e 6.
HTH

beat_gmat_09 wrote:

prachich1987 wrote:

beat_gmat_09 wrote: Number of ways such pattern i.e. BGBG can occur is selecting 2 out of 4= C(4,2) = 4!/2!*2! = 6
Required probability = 6*1/10 = 3/5

Thanks for the explanation.
The 1st method is clear.
But didn't understand the above part of 2nd method
can u plz explain a bit?

think of it as - in how many ways can you arrange 2 boys and 2 girls
which is - c(4,2) =6
or you can try to list the possibilities as-
bgbg
bbgg
gbgb
ggbb
bggb
gbbg
the value of conditional probability will be same in any particular combination, actually the probabilities of the events are added, but as the value of fraction remains same you can multiply the number of times you have to add I.e 6.
HTH

Thanks for the above
But I don't think its a conditional probability.
Here we are forming a team & hence order in which we choose a boy/girl doesn't matter
it wud hv mattered if we had to arrange these children in a row
Let me have our thoughts on this

prachich1987 wrote: But I don't think its a conditional probability.
Here we are forming a team & hence order in which we choose a boy/girl doesn't matter
it wud hv mattered if we had to arrange these children in a row
Let me have our thoughts on this

It is conditional (when the way it is solved), the first method is straight forward and the easiest.
Yes the order doesn't matter, but the number of possibilities in which you can form the team of 4 has to be considered by picking in some way, where order does not matter.
If the set is - B1B2G1G2, it doesn't matter if the first pick is B2 or B1 or G2 or G1, but you have to consider the number of ways as the probability is conditional.
By conditional it means 1st boy is selected - then next person selected is based on the condition that 1st person selected is a boy - so the selection set is constrained by it - the remaining set is - BGG (1st selected is boy), now next person to be selected can be a Boy or girl. If it is boy then probability is - 1 out of 3. If it were girl then probability is 2 out of 3 (higher than boy) - the first selection matters, in same way, 2nd selection matters to 3rd selection.
HTH.

beat_gmat_09 wrote:

prachich1987 wrote: But I don't think its a conditional probability.
Here we are forming a team & hence order in which we choose a boy/girl doesn't matter
it wud hv mattered if we had to arrange these children in a row
Let me have our thoughts on this

It is conditional (when the way it is solved), the first method is straight forward and the easiest.
Yes the order doesn't matter, but the number of possibilities in which you can form the team of 4 has to be considered by picking in some way, where order does not matter.
If the set is - B1B2G1G2, it doesn't matter if the first pick is B2 or B1 or G2 or G1, but you have to consider the number of ways as the probability is conditional.
By conditional it means 1st boy is selected - then next person selected is based on the condition that 1st person selected is a boy - so the selection set is constrained by it - the remaining set is - BGG (1st selected is boy), now next person to be selected can be a Boy or girl. If it is boy then probability is - 1 out of 3. If it were girl then probability is 2 out of 3 (higher than boy) - the first selection matters, in same way, 2nd selection matters to 3rd selection.
HTH.

Thanks!!!