Answer:
Yes, it is higher if w ≥ 3.
Explanation:
Given,
Number of men = 6,
Number of women = w,
Ways of choosing 2 men = C(6,2),
There can be two cases,
Case 1 :
If w ≥ 3,
Then the number of ways of choosing 1 men and 1 women > Ways of choosing two men,
So, the probability of choosing 1 men and 1 women is greater than that of choosing two men,
For eg : If w = 8,
[tex]^8C_1\times ^6C_1>^6C_2[/tex]
[tex]\implies \frac{^8C_1\times ^6C_1}{^{14}C_2}>\frac{^6C_2}{^{14}C_2}[/tex]
Similarly if w = 3,
[tex]\implies \frac{^3C_1\times ^6C_1}{^{14}C_2}>\frac{^6C_2}{^{14}C_2}[/tex]
Case 2 :
If 0 < w < 3,
Then the number of ways of choosing 1 men and 1 women < Ways of choosing two men,
So, the probability of choosing 1 men and 1 women is less than that of choosing two men
eg :
[tex]\frac{^2C_1\times ^6C_2}{^{14}C_2}<\frac{^6C_2}{^{14}C_2}[/tex]
