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tyleradobas
4 days ago
Math
Senior High School
answered
What is the experimental probability that the sum is 5?
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LinQiunan6
Genius
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Note:
(i) The outcomes (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) are called doublets.
(ii) The pair (1, 2) and (2, 1) are different outcomes.
1. Two dice are rolled. Let A, B, C be the events of getting a sum of 2, a sum of 3 and a sum of 4
respectively. Then, show that
(i) A is a simple event
(ii) B and C are compound events
(iii) A and B are mutually exclusive
Solution:
Clearly, we have
A = {(1, 1)}, B = {(1, 2), (2, 1)} and C = {(1, 3), (3, 1), (2, 2)}.
(i) Since A consists of a single sample point, it is a simple event.
(ii) Since both B and C contain more than one sample point, each one of them is a compound event.
(iii) Since A ∩ B = ∅, A and B are mutually exclusive.
2. Two dice are rolled. A is the event that the sum of the numbers shown on the two dice is 5, and B is
the event that at least one of the dice shows up a 3.
Are the two events (i) mutually exclusive, (ii) exhaustive? Give arguments in support of your answer.
Solution:
When two dice are rolled, we have n(S) = (6 × 6) = 36.
Now, A = {(1, 4), (2, 3), (4, 1), (3, 2)}, and
B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1,3), (2, 3), (4, 3), (5, 3), (6, 3)}
(i) A ∩ B = {(2, 3), (3, 2)} ≠ ∅.
Hence, A and B are not mutually exclusive.
(ii) Also, A ∪ B ≠ S.
Therefore, A and B are not exhaustive events.
(xi) getting a prime number as the sum:
Let E11 = event of getting a prime number as the sum. The events of a prime number as the sum will be E11 = [(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)] = 15
Therefore, probability of getting ‘a prime number as the sum’
Number of favorable outcomes
P(E11) = Total number of possible outcome
= 15/36
= 5/12
(xii) getting a doublet of even numbers:
Let E12 = event of getting a doublet of even numbers. The events of a doublet of even numbers will be E12 = [(2, 2), (4, 4), (6, 6)] = 3
Therefore, probability of getting ‘a doublet of even numbers’
Number of favorable outcomes
P(E12) = Total number of possible outcome
= 3/36
= 1/12
(xiii) getting a multiple of 2 on one die and a multiple of 3 on the other die:
Let E13 = event of getting a multiple of 2 on one die and a multiple of 3 on the other die. The events of a multiple of 2 on one die and a multiple of 3 on the other die will be E13 = [(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)] = 11
Therefore, probability of getting ‘a multiple of 2 on one die and a multiple of 3 on the other die’
Number of favorable outcomes
P(E13) = Total number of possible outcome
= 11/36
4. Two dice are thrown. Find (i) the odds in favour of getting the sum 5, and (ii) the odds against getting
the sum 6.
Solution:
We know that in a single thrown of two die, the total number of possible outcomes is (6 × 6) = 36.
Let S be the sample space. Then, n(S) = 36.
(i) the odds in favour of getting the sum 5:
Let E1 be the event of getting the sum 5. Then,
E1 = {(1, 4), (2, 3), (3, 2), (4, 1)}
⇒ P(E1) = 4
Therefore, P(E1) = n(E1)/n(S) = 4/36 = 1/9
⇒ odds in favour of E1 = P(E1)/[1 – P(E1)] = (1/9)/(1 – 1/9) = 1/8.
(ii) the odds against getting the sum 6:
Let E2 be the event of getting the sum 6. Then,
E2 = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
⇒ P(E2) = 5
Therefore, P(E2) = n(E2)/n(S) = 5/36
⇒ odds against E2 = [1 – P(E2)]/P(E2) = (1 – 5/36)/(5/36) = 31/5.
5. Two dice, one blue and one orange, are rolled simultaneously. Find the probability of getting
(i) equal numbers on both
(ii) two numbers appearing on them whose sum is 9.
Solution:
The possible outcomes are
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
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