Answer: The required probabilities are
[tex](a)~0\\\\(b)\dfrac{1}{9},\\\\(c)~\dfrac{1}{36}.[/tex]
Step-by-step explanation: Given that a pair of fair dice is rolled.
We are to find the probability of getting
(a) getting a sum of 1.
b) getting a sum of 5.
c) getting a sum of 12.
Let S be the sample space for the experiment of rolling a pair of fair dice.
Then, S = {(1,1), (1,2), (1,3), (1, 4), (1,5), (1,6), . . . , (6,5), (6,6)}.
And, n(S) =36.
(a) Let E denote the event of getting a sum of 1.
Since the sum of the numbers on two dice is minimum 2, so
E = { } ⇒ n(E) = 0.
Therefore, the probability of event E is
[tex]P(E)=\dfrac{n(E)}{n(S)}=\dfrac{0}{36}=0.[/tex]
(b) Let F denote the event of getting a sum of 5.
Then,
F = {(1,4), (2,3), (3,2), (4,1)} ⇒ n(F) = 4.
Therefore, the probability of event F is
[tex]P(F)=\dfrac{n(F)}{n(S)}=\dfrac{4}{36}=\dfrac{1}{9}.[/tex]
(c) Let G denote the event of getting a sum of 12.
Then,
G = {(6,6)} ⇒ n(G) = 1.
Therefore, the probability of event G is
[tex]P(G)=\dfrac{n(G)}{n(S)}=\dfrac{1}{36}.[/tex]
Thus, the required probabilities are
[tex](a)~0\\\\(b)\dfrac{1}{9},\\\\(c)~\dfrac{1}{36}.[/tex]
The probability of getting a sum of 1, the sum of 5, and the sum of 12 are 0, 1/9, and 1/36 respectively
Probability is the likelihood or chance that an event will occur.
For a pair of rolled dice, the total outcome will be 6² = 36
Probability = Expected outcome/Total outcome
a) Pr(getting a sum of 1) = 0/36 = 0
Note that since we have a pair of dice, the least sum we can have is 2
b) The event for getting a sum of 5 are (1, 4), (4, 1), (3, 2), (2, 3)
n(E) = 4
Pr( getting a sum of 5) = 4/36 = 1/9
c) The event for getting a sum of 12 are (6, 6)
n(E) = 1
Pr( getting a sum of 12) = 1/36
Hence the probability of getting a sum of 1, the sum of 5, and the sum of 12 are 0, 1/9, and 1/36 respectively
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