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 =)
3.5-total number of ways in which 4 balls can be selected out of 6 white and...
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