[3 marks sum]

Question 13 Marks

Find the probability that leap year selected at random, will contain 53 Sundays.

Answer

In a leap there are 366 days. In 366 days, we have 52 weeks and 2 days, Thus we can say that leap year has always 52 sundays.

The remaining two days can be
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
From above it is clear that there are 7 elementary events associated with this random experiment.
Clearly, the event A will happen if the last two days of the leap year are either Sunday and Monday or Saturday and Sunday.
n(E) = 2
P(E) = 2
$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{2}{7}$.

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Question 23 Marks

Two dice are thrown simultaneously. Find the probability of getting: a multiple of 3 as the sum.

Answer

n(s) = 36 i.e.
(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)}
Event = {multiple of 3 as a sum}
= {(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (3, 6), (6, 3), (5, 4), (4, 5), (6, 6)}
n(E) = 12
p(E) = ?
∴ P(E) = $\frac{ n ( E )}{ n ( S )}=\frac{12}{36}=\frac{1}{3}$

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Question 33 Marks

Two dice are thrown simultaneously. Find the probability of getting: a doublet of even number.

Answer

n(s) = 36 i.e.
(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)}
Event = { getting a doublet of even number }
Event = {(2, 2), (4,4), (6, 6)}
n(E) = 3
P(E) = ?
$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{3}{36}=\frac{1}{12}$

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Question 43 Marks

Two dice are rolled together. Find the probability of getting: a total of at least 10

Answer

n(s) = 36 i.e.
(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)
In throwing a dice, total possible outcomes={1,2,3,4,5,6}
n(s)=6
for two dice, n(s)= 6 x 6=36
E= event of getting a total of at least 10={(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)}
n(E)=6
Probability of getting a total of at least 10 = $\frac{n(E)}{n(s)}=\frac{6}{36}=\frac{1}{6}$

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Question 53 Marks

Two dice are thrown simultaneously. Find the probability of getting: the sum as a prime number

Answer

n(s) = 36 i.e.
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
n(s) = 36 i.e.
(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)
Event = { sum as a prime number }
i.e., = {(1, 1), (1, 2), (2, 1), (1, 4), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (6, 5) and (5, 6)}
n(E) = 15
P(E) = ?
∴ P(E) $\frac{ n ( E )}{ n ( s )}=\frac{15}{36}=\frac{5}{12}$

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Question 63 Marks

Two dice are thrown simultaneously. Find the probability of getting: an even number as the sum.

Answer

n(s) = 36 i.e.
(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)
Event = { even number as the sum }
i.e., = {(1, 1), (1, 3), (3, 1), (2, 2), (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (2, 6), (6, 2), (4, 4), (5, 3), (3, 5), (5, 5), (6, 4), (4, 6), and (6, 6)}
n( E) = 18
P(E) = ?
∴ P(E) = $\frac{ n ( E )}{ n ( S )}=\frac{18}{36}=\frac{1}{2}$

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Mathematics [3 marks sum]

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