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A CAR dealer has 106 vehiclels of which 48 are new,42 are trucks,and32 are red.if the are 6 used,red trucks and 12 new red vehicles,what is the probability of choosing a new or red vehicle?

2007-03-05 03:52:13 · 5 answers · asked by rose_west1 1 in Science & Mathematics Mathematics

5 answers

Mathematical representation of the question
n=106
Prob(new)=48/106
Prob(truck)=42/106
Prob(red)=32/106

Prob(used AND red AND trucks)=6/106
Prob(new AND red)=12/106

Solution:
Prob(new OR red)=Prob(new) + Prob(red) - Prob(new AND red)
=48/106 + 32/106 - 12/106

2007-03-05 04:08:00 · answer #1 · answered by Iqbal 1 · 0 0

i think my argument could not be properly gained. I say the prospect is 50% permit a ??, permit b ?? and randomly pick the values for a and b. As already pronounced, for a ? 0, P( a < b²) = a million, it particularly is trivial. in basic terms particularly much less trivial is the thought that P(a < 0 ) = a million/2 and subsequently P( a < b² | a ? 0) = a million and P( a < b² ) ? a million/2 Now evaluate what happens while a > 0 For a > 0, collectively because it is easy to coach there's a non 0 probability for a finite b, the shrink, the prospect is 0. a < b² is such as asserting 0 < a < b², bear in suggestions we are in basic terms watching a > 0. If this a finite era on an infinite line. The probability that a is an ingredient of this era is 0. P( a < b² | a > 0) = 0 As such we've a entire probability P( a < b² ) = P( a < b² | a ? 0) * P(a ? 0) + P( a < b² | a > 0) * P(a > 0) = a million * a million/2 + 0 * a million/2 = a million/2 bear in suggestions, it particularly is because of the countless instruments. whatever variety of era you draw on paper or on a working laptop or workstation you will detect a finite probability that seems to attitude a million. yet it particularly is as a results of the finite random selection turbines on the workstation and if we had this question asked with finite values there may well be a a answer extra advantageous than 50%. i don't mean to be condescending, yet please clarify why making use of the Gaussian to approximate a uniform distribution is a sturdy theory? are not countless numbers relaxing. Cantor while mad working with them! :)

2016-12-18 06:07:37 · answer #2 · answered by ? 4 · 0 0

48/106 + 32/106 -12/106 = 68/106 = 34/53

2007-03-05 03:58:29 · answer #3 · answered by ag_iitkgp 7 · 0 0

P(N) = 48/106
P(R) = 32/106
P(N int r) = 12/106
->P(N union R) = P(N) + P(R) - P(N int R)
= 48/106 + 32/106 - 12/106
= 68/106
= 34/53

2007-03-05 04:14:21 · answer #4 · answered by ganesan 2 · 0 0

106 - (42 - 32) = __
106 - 10 = 96
= 96/ 106
= 48/ 53
Fraction Form: 48/ 53, Percent Form: 90.56603774, Ratio: 48 to 53, Ratio: 48:53, Ratio: 48/53.

2007-03-05 04:12:36 · answer #5 · answered by Anonymous · 0 1

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