An accounting firm has recently recruited six graduates: three men and three women. Three of the graduates are to be selected at random to work in the firm’s suburban office. Construct the probability tree. What is the probability that at least two women will be selected? Round your answer to three decimal places.

a) The probability tree is constructed below

b) the probability that at least two women will be selected is 0.500

Step-by-step explanation:

Given the data in the question;

a) The probability tree is as follows;

                                                 3 men, 3 women

                                                            ↓

           |

            ↓                             ↓                                  ↓                               ↓

         3 men                    2 men                          1 man                        0 men

      0 women                1 woman                     2 woman                   3 women

b) the probability that at least two women will be selected

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

so

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

Therefore, the probability that at least two women will be selected is 0.500

30.5*10=350$

you have to multiply the number without the decimal point then add it later but if its tenn u must move the place value of the decimal point 1 to the right.