c) Let B = event of getting 1 head and 1 tail. Then,
B = {HT, TH } and, therefore, n(B) = 2.
Therefore, P(getting 1 head and 1 tail)
= P(B)
= n(B)/n(S)
= 2/4
= 1/2.
2. A pair of dice is rolled
a) S = {(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)}
b) Same number on two faces: Let A = event of getting a same number on two face.
The number which A will be A = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] = 6
Therefore, probability of getting ‘a same number on two faces’
Number of favorable outcomes
P(A) = Total number of possible outcome
= 6/36
= 1/6
c) Let B = event of getting a sum of 7. The number which is a sum of 7 will be B = [(6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)] = 6
Therefore, probability of getting ‘a sum of 7’
Number of favorable outcomes
P(B) = Total number of possible outcome
= 6/36
= 1/6
d) Let C = event of getting a sum of 10.
The number which is a sum of 10 will be C = [(6, 4), (5, 4), (4, 6)] = 3
Therefore, probability of getting ‘a sum of 10’
Number of favorable outcomes
P(C) = Total number of possible outcome
= 3/36
= 1/12
e) Let D = event of getting both even numbers.
The number which is are both even numbers will be D = [(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)] = 9
Therefore, probability of getting ‘both even numbers’
Number of favorable outcomes
P(D) = Total number of possible outcome
= 9/36
= 1/4
f) Let E = event of getting one number is 3 and one is even.
The number which is one number is 3 and one is even will be E = [(3, 2), (3, 4), (3, 6)] = 3
Therefore, probability of getting ‘one number is 3 and one is even’
Number of favorable outcomes
P(E) = Total number of possible outcome
= 3/36
= 1/12
g) Let F = event of getting a sum is less than five
The number which is a sum is less than five will be F = [(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)] = 10
Therefore, probability of getting ‘a sum is less than five’
Number of favorable outcomes
P(F) = Total number of possible outcome
= 10/36
= 5/18
Activity 5
1) Let A = event of getting 2 heads. Then,
A = {HH} and, therefore, n(A) = 1.
Therefore, P(getting 2 heads)
= P(A)
= n(A)/n(S)
= 1/4.
2)
a) Let A = event of getting both odd numbers
The number which both odd numbers will be A = [(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)] = 9
Therefore, probability of getting ‘both odd numbers’
Number of favorable outcomes
P(A) = Total number of possible outcome
= 9/36
= 1/4
b) Let B = event of getting a sum of more than 10.
The number which is a a sum of more than 10 will be B = [(6, 4), (5, 5), (4, 6), (6, 5), (5, 6), (6, 6)] = 6
Therefore, probability of getting ‘a sum of more than 10’
Number of favorable outcomes
P(B) = Total number of possible outcome
= 6/36
= 1/6
3)
a) Red card:
Cards of hearts and diamonds are red cards.
Number of hearts = 13
Number of diamonds = 13
Therefore, total number of red cards out of 52 cards = 13 + 13 = 26
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Verified answer
Answer:
Activity 4
1. Two coins are tossed
a) S = {HH, TT, HT, TL|}
b) Let A = event of getting 2 tails. Then,
A = {HH} and, therefore, n(A) = 1.
Therefore, P(getting 2 faces)
= P(A)
= n(A)/n(S)
= 1/4.
c) Let B = event of getting 1 head and 1 tail. Then,
B = {HT, TH } and, therefore, n(B) = 2.
Therefore, P(getting 1 head and 1 tail)
= P(B)
= n(B)/n(S)
= 2/4
= 1/2.
2. A pair of dice is rolled
a) S = {(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)}
b) Same number on two faces: Let A = event of getting a same number on two face.
The number which A will be A = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] = 6
Therefore, probability of getting ‘a same number on two faces’
Number of favorable outcomes
P(A) = Total number of possible outcome
= 6/36
= 1/6
c) Let B = event of getting a sum of 7. The number which is a sum of 7 will be B = [(6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)] = 6
Therefore, probability of getting ‘a sum of 7’
Number of favorable outcomes
P(B) = Total number of possible outcome
= 6/36
= 1/6
d) Let C = event of getting a sum of 10.
The number which is a sum of 10 will be C = [(6, 4), (5, 4), (4, 6)] = 3
Therefore, probability of getting ‘a sum of 10’
Number of favorable outcomes
P(C) = Total number of possible outcome
= 3/36
= 1/12
e) Let D = event of getting both even numbers.
The number which is are both even numbers will be D = [(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)] = 9
Therefore, probability of getting ‘both even numbers’
Number of favorable outcomes
P(D) = Total number of possible outcome
= 9/36
= 1/4
f) Let E = event of getting one number is 3 and one is even.
The number which is one number is 3 and one is even will be E = [(3, 2), (3, 4), (3, 6)] = 3
Therefore, probability of getting ‘one number is 3 and one is even’
Number of favorable outcomes
P(E) = Total number of possible outcome
= 3/36
= 1/12
g) Let F = event of getting a sum is less than five
The number which is a sum is less than five will be F = [(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)] = 10
Therefore, probability of getting ‘a sum is less than five’
Number of favorable outcomes
P(F) = Total number of possible outcome
= 10/36
= 5/18
Activity 5
1) Let A = event of getting 2 heads. Then,
A = {HH} and, therefore, n(A) = 1.
Therefore, P(getting 2 heads)
= P(A)
= n(A)/n(S)
= 1/4.
2)
a) Let A = event of getting both odd numbers
The number which both odd numbers will be A = [(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)] = 9
Therefore, probability of getting ‘both odd numbers’
Number of favorable outcomes
P(A) = Total number of possible outcome
= 9/36
= 1/4
b) Let B = event of getting a sum of more than 10.
The number which is a a sum of more than 10 will be B = [(6, 4), (5, 5), (4, 6), (6, 5), (5, 6), (6, 6)] = 6
Therefore, probability of getting ‘a sum of more than 10’
Number of favorable outcomes
P(B) = Total number of possible outcome
= 6/36
= 1/6
3)
a) Red card:
Cards of hearts and diamonds are red cards.
Number of hearts = 13
Number of diamonds = 13
Therefore, total number of red cards out of 52 cards = 13 + 13 = 26
Therefore, probability of getting ‘a red card’
Number of favorable outcomes
P(H) = Total number of possible outcome
= 26/52
= 1/2
b) Ace card
Number of ace card out ofthe whole deck = 4
Therefore, probability of getting ‘an ace card’
Number of favorable outcomes
P(H) = Total number of possible outcome
= 4/52
= 1/13
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