Answer: 0.6
Step-by-step explanation:
Given
There are 3 women and 3 men available to form a 2 member committee.
No of ways, a man can be chosen are [tex]3[/tex]
Similarly, no of ways a woman can be chosen are 3
Therefore, probability of choosing 2 members i.e. 1 man and 1 women is
[tex]\Rightarrow P=\dfrac{3}{6}\times \dfrac{3}{5}[/tex]
If the order matter, then
[tex]\Rightarrow P=2\times \dfrac{3\times 3}{6\times 5}\\\\\Rightarrow P=0.6[/tex]
The probability that 1 woman and 1 man will be chosen is "0.6".
According to the question, the club has nominated:
Now,
→ The number of ways to select two persons from 6 will be:
= [tex]\binom{6}{2}[/tex]
hence,
The probability that one woman and one man will be chosen will be:
→ [tex]P(1 \ women \ and \ 1 \ women) = \frac{\binom{3}{1} \binom{3}{1}}{\binom{6}{2}}[/tex]
[tex]= 3\times \frac{3}{15}[/tex]
[tex]= \frac{9}{15}[/tex]
[tex]= 0.6[/tex]
Thus the above answer is right.
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