(a) A speaks truth 30% of the times, B speaks truth 80% of the times. What is the
probability that they will contradict each other? (2 Marks)
(b) In how many ways can a team of 11 cricketers be chosen from 9 batsman and
6 bowlers so that at-least 3 bowlers are included? (2 Marks)
(c) The Probability that there is at least one error in an account statement prepared
by A is 0.3 for B is 0.4 and for C it is 0.45.If A,B and C prepared 20,10 and 40
statements respectively .What is the expected number of correct statements. (2 Marks)
(d) A company requires 10, 15 and 20 persons for sales promotion in 3 zones A, B
and C respectively .45 people are available.
(i) In how many way can they be allocated? (1 Mark)
(ii) If out of 45 people, 3 particular persons are to be posted to A , 5 to B
and 2 to C in how many ways can it be done?
Answers
Given : A speaks truth 30% of the times, B speaks truth 80% of the times
To Find : What is the probability that they will contradict each other
Solution:
A speaks truth 30% of the times = 0.3
A does not speaks truth = 0.7
B speaks truth 80% of the times = 0.8
B does not speaks truth = 0.2
contradict each other = A - Truth B does not + A Does not B - Truth
= (0.3)(0.2) + (0.7)(0.8)
= 0.06 + 0.56
= 0.62
ways a team of 11 cricketers be chosen from 9 batsman and 6 bowlers so that at-least 3 bowlers are included
3 bowlers 8 Batsman + 4 bowlers 7 Batsman + 5 bowlers 6 Batsman + 6 bowlers 5 Batsman
= ⁶C₃.⁹C₈ + ⁶C₄.⁹C₇ + ⁶C₅.⁹C₆ + ⁶C₆.⁹C₅
= 180 + 540 + 504 + 126
= 1350
other way
9 batsman can not be selected as atleast 3 bowlers
Hence ¹⁵C₁₁ -⁹C₉.⁶C₂ = 1365 - 15 = 1350
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