Question

1500 families with 2 children were delected randomly and the following data were recorded:
If a family is chosen at random, compute the probability that it has:
(i) No girl
(ii) 1 girl
(iii) 2 girls
(iv) At most one girl
(v) More girls than boys.

Answer 1

Total numbers of families = 1500

Let E1, E2 , E3 , E4 and E5 be the events choosing no girl,  1 girl and 2girls ,at most one girl  and more girls than boys

(i) Numbers of families having no girl = 211

Probability ,P(E1) = Numbers of families having 0 girls / Total numbers of families = 211/1500

P(E1) = 211/1500

 

(ii) Numbers of families having 1 girls = 814

Probability ,P(E2)= Numbers of families having 1 girls/Total numbers of families = 814/1500 = 407/750

P(E2) = 407/750

 

(iii)  Numbers of families having 2 girls = 475

Probability P(E3) = Numbers of families having 2 girls/Total numbers of families = 475/1500 = 19/60

P(E3) = 19/60

(iv) Number of families having at most one girl  = 211 + 814 = 1025

Probability P(E4) = Numbers of families having at most one girl  /Total numbers of families = 1025/1500 = 205/300 = 41/60

P(E4) = 41/60

 

(v) Number of families having more girls than boys = 475

Probability P(E5) = Numbers of families having more girls than boys /Total numbers of families = 475/1500 = 19/60

P(E5) = 19/60

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Answer 2

Answer:

Total numbers of families = 1500.......n(s)

Let E1, E2 , E3 , E4 and E5 be the events choosing no girl,  1 girl and 2girls ,at most one girl  and more girls than boys.

(i) n(E1)=211 E1-No girl

P(E1) = n(E1)/ n(s) = 211/1500

P(E1) = 211/1500

 

(ii) n(E2) = 814 E2-1 girl

P(E2)= n(E2)/n(s) = 814/1500 = 407/750

P(E2) = 407/750

 

(iii) n(E3) = 475 E3-2 girls

P(E3) = n(E3)/n( s) = 475/1500 = 19/60

P(E3) = 19/60

(iv) n(E4)  = 211 + 814 = 1025 E4- At most one girl

P(E4) = n(E4)  /n(s) = 1025/1500 = 205/300 = 41/60

P(E4) = 41/60

 

(v) n(E5) = 475 E5-Numbers of families having more girls than boys

P(E5) = n(E5) /n(s)= 475/1500 = 19/60

P(E5) = 19/60