Question 2 1500 families with 2 children were selected randomly, and the following data were recorded:
Number of girls in a family 2 1 0
Number of families 475 814 211
Compute the probability of a family, chosen at random, having
(i) 2 girls
(ii) 1 girl
(iii) No girl
Also check whether the sum of these probabilities is 1.
Class 9 - Math - Probability Page 283
Answers
Probability is the study of the chances of events happening.
Event:
A Possible outcome or combination of outcomes is called event.
The probability of happening our event always lies from 0 to 1.
The sum of all the probabilities of all possible outcomes of an experiment is 1.
Required probability=number of trials in which the event E has happened/Total number of trials
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Solution:
Total numbers of families = 1500
Let E1, E2 & E3 be the events choosing 2girls , 1 girl and no girl.
(i) Numbers of families having 2 girls = 475
Probability P(E1)= Numbers of families having 2 girls/Total numbers of families
= 475/1500 = 19/60
P(E1)= 19/60
(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 no girl = 211
Probability ,P(E3)= Numbers of families having 0 girls/Total numbers of
families
= 211/1500
P(E3)= 211/1500
Sum of the probabilities= 19/60 + 407/750 + 211/1500
= (475 + 814 + 211)/1500
= 1500/1500 = 1
Yes, the sum of these probabilities is 1.
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Hope this will help you....
Let x ,x1, and x2 be the event of getting no girls ,1 girls ,2 girl.
1). P(x 2)== Probability of a family having 2 girls
=475/1500
=19/60
2)P(x1)== Probability of a family having 1 girl.
=814/1500
=407/705
3)P( x)=. { ""}No girls
=211/1500
So,sum of Probabilities
P(x)+P(x1)+P(x2 )
=211/1500+407/750+19/60
=211+814+475/1500
=1500/1500
=1ans.
Hope it is helpful ...