Question

3.5-total number of ways in which 4 balls can be selected out of 6 white and 9 black balls = 15P4 = 32760

Number of ways in which 4 balls can be selected where first 2 are white and last 2 black = 6P2*9P2 = 2160

Thus, required probability = 2160/32760 = 0.0659

3.16)

Let the event that baby survives delivery be denoted by B

and the event that C section is performed be denoted by C

P(B) = 0.98, P(C) = 0.15

P(B|C) = 0.96

Thus, P(C|B) = P(B|C)*P(C)/P(B)

= 0.96*0.15/0.98 = 0.14694

So P(C'|B) = 1 - P(C|B) = 0.85306

To find P(B|C') = P(C'|B)*P(B)/P(C')

= 0.85306*0.98/0.85

= 0.98353

3.19)

Percentage of male in original class =62

Thus, percentage of female in original class = 38

Percentage of original class who attended the party

= 0.48*38 + 0.37*62 = 41.18

Thus, percentage of women among those attending the party

= 0.48*38/41.18 * 100 = 44.29

(a) 44.29%

(b) 41.18%

These are questions and solutions ,MAY you please write down the answers either by hand or by typing it by laptop so I can submit it directly

I need it in 1 hour from now and thanks in advance =)

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Answer 1

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