An accounting firm has recently recruited six graduates: three men and three women. Three of the graduates are to be selected at

Answer:

a) The probability tree has been created below

b) The likelihood of selecting at least two women is 0.500

Step-by-step explanation:

Based on the information provided in the question;

a) The constructed probability tree is outlined as follows;

                                                 3 men, 3 women

                                                            ↓

           _______________________|________________________

            ↓                             ↓                                  ↓                               ↓

         3 men                    2 men                          1 man                        0 men

      0 women                1 woman             2 women           3 women

b) The probability of having at least two women selected

p( at least two women would be selected) = P( there are 2 women out of 3 ) + P( there are 3 women out of 3 )

Thus,

p( at least two women would be selected) = C

³C₂ × ³C₁ / ⁶C₃ + ³C₃³C₀ / ⁶C₃  

= 3!/(2!(3-2)!) × 3!/(1!(3-1)!)  / 6!/(3!(6-3)!) + 3!/(3!(3-3)!) × 3!/(0!(3-0)!) / 6!/3!(6-3)!)

= 3 × 3 / 20 + 1 ×  1 / 20

= 9/20 + 1/20

= 0.45 + 0.05

p( at least two women would be selected) = 0.500

Thus, the probability of selecting at least two women is 0.500