A group consist of 12 persons of which 3 are extremely patient other 6 are extremely honest and rest are extremely kind .A person from the group is selected at random assuming that each person is equally likely to be selected find the probability of selecting a person who is
1.extremely patient
2.extremely kind or honest
p lzz answer plzzz help me.....
Answers
Answered by
67
1.probability = no.of favourable outcome/ total observation
=3/12=1/4
2.probability= 6(kind)+3(honest)/12
=9/12=3/4
=3/12=1/4
2.probability= 6(kind)+3(honest)/12
=9/12=3/4
Answered by
43
Total no. of persons = 12
No. of extremely patient persons = 3
No. of extremely honest persons = 6
No. of extremely kind persons = 12 – 9 = 3
(i) Required probability = 3/12 = 1/4.
(ii) Required probability = 6+3/12 = 9/12 = 3/4 Honesty.
No. of extremely patient persons = 3
No. of extremely honest persons = 6
No. of extremely kind persons = 12 – 9 = 3
(i) Required probability = 3/12 = 1/4.
(ii) Required probability = 6+3/12 = 9/12 = 3/4 Honesty.
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