A red ball can be drawn in two mutually exclusive ways
(i) Selecting bag I and then drawing a red ball from it.
(ii) Selecting bag II and then drawing a red ball from it.
Let E1, E2 and A denote the events defined as follows:
E1 = selecting bag I,
E2 = selecting bag II
A = drawing a red ball
Since one of the two bags is selected randomly, therefore
P(E1) = 1/2 and P(E2) = 1/2
Now, = Probability of drawing a red ball when the first bag has been selected = 4/7
= Probability of drawing a red ball when the second bag has been selected = 2/6
Using the law of total probability, we have
P(red ball) = P(A) =
=
n(S) = 6, n(E) = (4, 6) = 2
? P(E) = 2/6 = 1/3
Here, A = ? 8800, T =2 yr, R = 5%
We know
SI = ART/(100 + RT) = (8800 x 5 x 2) / (100 + 5 x 2)
= (8800 x 10) / 110
= ? 800
3rd term = (2nd term) x 3 - 4 = 26 x 3 - 4 = 74.
4th term = (3th term) x 3 - 4 = 74 x 3 - 4 = 218.
5th term = (4th term) x 3 - 4 = 218 x 3 - 4 = 650.
∴ 5th term must be 650 instead of 654.
Other side = ?5 2 - 42
= ? 9
=3 m
So The area of the rectangular field = 4 x 3
= 12 m2
0.27 = (27 - 2)/90 = 25/90 = 5/18
For 50 students, food is sufficient for 45 days
? For 1 student, food is sufficient for 45 x 50 days
and for 75 students, food is sufficient for (45 x 50)/75 days. i,e., for 30 days.
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