varundixitmro2512 wrote:
The total no of ways we can arrange 3 men and 2 women(MMMWW) is 5!/3!2!=10 as we have 3 males and 2 women.
Now there is only 1 way when two men cant be consecutive i.e. MWMWM
Therefore the prob atleast 2 men will be consecutive will be 9/10.
Though I marked the wrong answer, but after some thinking I got with the explanation.
Yes this is another plausible method w/r/t a fundamental statistical formula - # desired outcome(s)/ # total outcome (s)
We must first calculate the probability of an arrangement in which none of the boys appear consecutively or more accurately the number of arrangements in which boys do not appear consecutively over the total number of arrangements possible
Total number of arrangement possible= 5! because there are 5 elements and 5 slots
Now, the number of arrangement in which boys do not appear consecutively follows the pattern
BGBGB - though this does not actually mean the number of outcomes, or the number of occurrences, in which boys do not appear consecutively is 1 single possibility because all three boys are still distinct entities (e.g if you have three boys fred, bob, and sam then you could have (f)g(b)g(s) or (b)g(s)g(f) - etc) so the actual number of ways we can arrange the three boys with respect to that pattern is 3! More fundamentally, this question is a permutation so like wise... the number of ways we can arrange girls in that pattern is 2! . Hence
1- 3!2!/5!
Hence
9/10