Answer:
a) P(A) = 0,4607 or P(A) = 46,07 %
b) P(B) = 0,7120 or 71,2 %
c) P(C) = 0,2055 or P(C) = 20,55 %
Step-by-step explanation:
We will use two concepts in solving this problem.
1.- The probability of an event (A) is for definition:
P(A) = Number of favorable events/ Total number of events FE/TE
2.- If A and B are complementary events ( the sum of them is equal to 1) then:
P(A) = 1 - P(B)
a) The total number of events is:
C ( 24,4) = 24! / 4! ( 24 - 4 )! ⇒ C ( 24,4) = 24! / 4! * 20!
C ( 24,4) = 24*23*22*21*20! / 4! * 20!
C ( 24,4) = 24*23*22*21/4*3*2
C ( 24,4) = 24*23*22*21/4*3*2 ⇒ C ( 24,4) = 10626
TE = 10626
Splitting the group of tanks in two 6 with h-v and 24-6 (18) without h-v
we get that total number of favorable events is the product of:
FE = 6* C ( 18, 3) = 6 * 18! / 3!*15! = 18*17*16*15!/15!
FE = 4896
Then P(A) ( 1 tank in the sample contains h-v material is:
P(A) = 4896/10626
P(A) = 0,4607 or P(A) = 46,07 %
b) P(B) will be the probability of at least 1 tank contains h-v
P(B) = 1 - P ( no one tank with h-v)
Again Total number of events is 10626
The total number of favorable events for the ocurrence of P is C (18,4)
FE = C (18,4) = 18! / 14!*4! = 18*17*16*15*14!/14!*4!
FE = 18*17*16*15/4*3*2 = 3060
Then P = 3060/10626
P = 0,2879
And the probability we are looking for is
P(B) = 1 - 0,2879
P(B) = 0,7120 or 71,2 %
c) We call P(C) the probability of finding exactly 1 tank with h-v and t-i
having 4 with t-i tanks is:
reasoning the same way but now having 4 with t-i (impurities) number of favorable events is:
FE = 6*4* C(14,2) = 24 * 14!/12!*2!
FE = 24* 14*13*12! / 12!*2
FE = 24*14*13/2 ⇒ FE = 2184
And again as the TE = 10626
P(C) = 2184/ 10626
P(C) = 0,2055 or P(C) = 20,55 %
