The probability that a student passes a physics test is \(\frac{2}{3}\). If he takes three physics tests, what is the probability that he passes two of the tests?
\(\frac{2}{27}\)
\(\frac{3}{27}\)
\(\frac{4}{27}\)
\(\frac{5}{3}\)
Explanation
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Discussions (6)
hello myschool , the answer isnt even here. and the answer is definitely not C.
Pass(P) - 2/3
fail(F)-1/3
Now, the pr (he passes two test) is
(PPF) + (PFP) + (FPP) same as 3x(PPF)
NOTE: There wasnt any exact order in the question, so its possibke that he fails the 1st test and passes the next two OR pass the 1st test fails the 2nd and passes the 3rd OR passes thele first 2 tests and fails the third.
what you did as your explanation myschool means that the question asked you to find the pr (that the boys passes tge first two test and fails the third)
1st method
so the correct answer is 3(2/3 x 2/3 x 1/3) = 4/9
2nd method
(2/3 x 2/3 x1/3) + (2/3 x 1/3 x 2/3) + (1/3 x 2/3 x 2/3) =4/9
3rd Method
Using binomial distribution.
P = 2/3
q = 1/3
n = 3
r = 2
T = nCr . P^r . q^(n - r)
T = 3C2 . (2/3)^2. (1/3)^1
T = 4/9
Wrong answer
The probability of passing one of the 3 tests is either PPF OR PFP OR FPP
That is: (2/3*2/3*1/3) + (2/3*1/3*2/3) + (1/3*2/3*2/3)
3*4/27 = 4/9
