1500 families with two children was selected recording the and the following data were recorded
No. of girls in family 2. 1. 0
No. of families. 475. 814 211
compute the probability of a family student at random having
I) 2 girls
II) At least 1 girl
III) No girl
IV) Almost 1 girl
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Answer:
1) p(g2) = 2/475 = 19/60
2)p(g1) = 1/814 = 407/750
3) p(Ng) = 211/1500
4)p(g2)+p(g1)+p(Ng)=
455/1500+814/1500+211/1500
1500/1500= 1
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