3 |
4 |
4 |
7 |
1 |
8 |
3 |
7 |
4 |
7 |
Number of white balls = 8.
P (drawing a white ball) = | 8 | = | 4 | . |
14 | 7 |
1 |
3 |
3 |
4 |
7 |
19 |
8 |
21 |
9 |
21 |
1 |
3 |
Let E | = event that the ball drawn is neither red nor green |
= event that the ball drawn is blue. |
∴ n(E) = 7.
∴ P(E) = | n(E) | = | 7 | = | 1 | . |
n(S) | 21 | 3 |
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
∴ P(E) = | n(E) | = | 7 | . |
n(S) | 8 |
10 |
21 |
11 |
21 |
2 |
7 |
5 |
7 |
10 |
21 |
Let S be the sample space.
Then, n(S) | = Number of ways of drawing 2 balls out of 7 | |||
= 7C2 ` | ||||
|
||||
= 21. |
Let E = Event of drawing 2 balls, none of which is blue.
∴ n(E) | = Number of ways of drawing 2 balls out of (2 + 3) balls. | |||
= 5C2 | ||||
|
||||
= 10. |
∴ P(E) = | n(E) | = | 10 | . |
n(S) | 21 |
1 |
10 |
2 |
5 |
2 |
7 |
5 |
7 |
2 |
7 |
P (getting a prize) = | 10 | = | 10 | = | 2 | . |
(10 + 25) | 35 | 7 |
S = {1, 2, 3, ......16}
E = {2, 3, 5, 7, 11, 13}
? P(E) = n(E)/n(S) = 6/16= 3/8
P (red) = 9 / (9 + 7 + 4) = 9/20
? P(not-red) = (1 - 9/20) = 11/20
P (getting a prize ) = 20 / (20 + 15 ) = 20 / 35 = 4/7
Number of cases favourable of E = 3
Total Number of cases = (3 + 5 ) = 8
? P(E) = 3/8
Probability of getting head in one trail = 1/2
? Reqd Probability of getting heads in both the trails = 1/2 x 1/2 = 1/4
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