A box containing 12 balls numbered 1 to 12. If two balls are taken at the same time at random, the odds of getting an odd numbered ball are
Answers
A box containing 12 balls numbered 1 to 12. If two balls are taken at the same time at random, the odds of getting an odd numbered ball are 5/22.
DISCUSSION :
Chance of an event
Suppose there is an experiment having a sample space S and K is one of the events of the experiment with:
n (S) = many members of the sample room
n (K) = many members of the event K
The chance for occurrence of K is defined as follows:
P (K) = n (K) / n (S)
↓↓↓↓↓
Known :
12 balls with numbers 1 to 12
Asked: The chance for an odd numbered ball to be taken if two balls are randomly drawn?
Answer:
There are 12 balls numbered 1 to 12, so many sample spaces:
n (S) = 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
n (S) = (n / 2) (a + Un)
n (S) = (11/2) (11 + 1)
n (S) = (11/2) (12)
n (S) = 11. 6
n (S) = 66
Taken both balls with odd numbers
= {1, 3, 5, 7, 9, 11}
= there are 6 balls, so
n (A) = many members of event A
n (A) = 5 + 4 + 3 + 2 + 1
n (A) = 15
That is :
(1, 3), (1, 5), (1, 7), (1, 9), (1, 11) = 5
(3, 5), (3, 7), (3, 9), (3, 11) = 4
(5, 7), (5, 9), (5, 11) = 3
(7, 9), (7, 11) = 2
(9, 11) = 1
The chances of getting both odd numbered balls are:
P (A) = n (A) / n (S)
P (A) = 15/66
P (A) = 5/22
Step-by-step explanation:
There are 12 balls numbered 1 to 12, so many sample spaces:
n (S) = 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
n (S) = (n / 2) (a + Un)
n (S) = (11/2) (11 + 1)
n (S) = (11/2) (12)
n (S) = 11. 6
n (S) = 66
Taken both balls with odd numbers
= {1, 3, 5, 7, 9, 11}
= there are 6 balls, so
n (A) = many members of event A
n (A) = 5 + 4 + 3 + 2 + 1
n (A) = 15
That is :
(1, 3), (1, 5), (1, 7), (1, 9), (1, 11) = 5
(3, 5), (3, 7), (3, 9), (3, 11) = 4
(5, 7), (5, 9), (5, 11) = 3
(7, 9), (7, 11) = 2
(9, 11) = 1
The chances of getting both odd numbered balls are:
P (A) = n (A) / n (S)
P (A) = 15/66
P (A) = 5/22