number of women --- x
number of men ---- 3x
total = 4x
so prob(choosing a woman) = x/4x
prob(choosing a man) = 3x/4x
prob(choosing 3 men of 3) = C(3,3)(3x/4x)^3 (x/4x)^0 = 1(27/64)(1) = 27/64
or
prob(choosing 3 men) = (3x/4x)( (3x-1)/(4x-1) )( (3x-2)/(4x-2) ) = 3x(3x-1)(3x-2) / (4x(4x-1)(4x-2) )
= (27x^3 - 27x^2 + 6x)/(64x^3 - 48x^2 + 8x)
now we want the
lim [ (27x^3 - 27x^2 + 6x)/(64x^3 - 48x^2 + 8x) ] as x ---->∞
= 27/64
2 women, 1 man:
= C(3,2) (1/4)^2 (3/4)^1 = 3(1/16)(3/4) = 9/64
or, as before
prob = (x/4x)*(x-1)/(4x-1)*(3x)/4x-2) + (3x)/4x-2)*(x/4x)*(x-1)/(4x-1) + (x/4x)*(3x)/4x-2)*(x-1)/(4x-1)
prob = Lim [3(x)(x-1)(3x)/(4x(4x-1)(4x-2) ) as x ---->∞
= lim (9x^3 - 9x^2)/(64x^3 - 48x^2 + 8x) as x ---->∞
= 9/64
Maths
In a large crowd,there are three times as many men as women.Three people are chosen at random.Assuming that there are so many people that choosing three has a negligible effect on the proportion of men to women,find the probability that they are (a) all men (b) 2 women and 1 man
1 answer