How many five-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7, and 8? repeated digits are allowed?
1 answer:
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Answer:
(a) 185/703
(b) 518/703
(c) 45/703
(d) 140/703
(e) 140/703
(f) 378/703
(g) 185/703
Step-by-step explanation:
Let the event of pledge member be P and event of full member be F
(a)If the first is a pledge, then the second is either a pledge or a member.
required Probabability = [P n (PUF)]
= Pr(PnP) U Pr (PnF)
= 10/38 × 9/37 + 10/38 × 28/38
= 90/1406 + 289/1406 = 370/1406
= 185/703
(b) if the person selected is not a pledge then the person is a full member and the second is either a full member or pledge
The required probability = Fn(FUP) =Pr (FnP) U Pr(FnF)
= 28/38 × 10/37 + 28/38 × 27/37
= 280/1406 + 756/140 = 1036/1406
= 518/703
(c) probability required= Pr(P/P) = Pr(PnP) = 10/38 × 9/37
= 90/1406
= 45/703
(d) Probability required = Pr(F/P)
= Pr(P) nPr(F) = 10/38 × 28/37
= 280/1406
= 140/703
(e) Probability required = Pr(P/F)
= Pr(F) n Pr(P)
= 28/38 × 10/37 =280/1406
=140/704
(f) Probability required = Pr (FnF)
= Pr(F) n Pr(F)
= 28/38 × 27/37= 756/1406
= 378/703
(g) it is the either the first person is a pledge member or a full member
Probability required = Pr[(PnP) U (FnP)]
= Pr(PnP) + Pr(FnP)
= 10/38 × 9/37 + 28/38 × 10/37
= 90/1406 + 280/1406
= 370/1406
= 185/703
(h) see attachment.
