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junior college 1 | H3 Maths
2 Answers Below
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Ion, pls help me
Date Posted: 5 years ago
Views: 412
a)
Case 1: number starts with zero but does not end with zero.
0 _ _ _
Number of ways to choose 2nd digit = 10
Number of ways to choose 3rd digit = 10
Number of ways to choose 4th digit= 9
Total for Case 1: = 10 x 10 x 9 = 900
Case 2: number ends with zero but does not start with zero.
_ _ _ 0
Number of ways to choose 1st digit = 9
Number of ways to choose 2nd digit = 10.
Number of ways to choose third digit = 10
Total for Case 2 = 9 x 10 x 10 = 900
Case 3: Both 1st and 4th digit are 0.
0 _ _ 0
Number of ways to choose 2nd digit
= 10.
Number of ways to choose 3rd digit
= 10.
Total for Case 3 = 10 x 10 = 100
Total of Case 1 , 2 and 3
= 900 + 900 + 100 = 1900
From 0000 to 9999, there are 10000 numbers.
So, probability = 1900 ÷ 10000 = 0.19
Case 1: number starts with zero but does not end with zero.
0 _ _ _
Number of ways to choose 2nd digit = 10
Number of ways to choose 3rd digit = 10
Number of ways to choose 4th digit= 9
Total for Case 1: = 10 x 10 x 9 = 900
Case 2: number ends with zero but does not start with zero.
_ _ _ 0
Number of ways to choose 1st digit = 9
Number of ways to choose 2nd digit = 10.
Number of ways to choose third digit = 10
Total for Case 2 = 9 x 10 x 10 = 900
Case 3: Both 1st and 4th digit are 0.
0 _ _ 0
Number of ways to choose 2nd digit
= 10.
Number of ways to choose 3rd digit
= 10.
Total for Case 3 = 10 x 10 = 100
Total of Case 1 , 2 and 3
= 900 + 900 + 100 = 1900
From 0000 to 9999, there are 10000 numbers.
So, probability = 1900 ÷ 10000 = 0.19
tq
b)
Number containing exactly two non-zero
digits means two digits are zero and two digits are not zero.
6 possible cases:
0 0 _ _
0 _ 0 _
0 _ _ 0
_ 0 _ 0
_ 0 0 _
_ _ 0 0
Or, recognise that there are two identical zeroes. We need to arrange these zeroes.
So number of ways to arrange
= 4 C 2 = 6
For each case, there are 9 ways to choose each remaining digit (1-9 only, has to be two non-zero digits)
So 9 ways for one digit x 9 ways for the other digit x 6 cases = 486.
Provability = 486 ÷ 10000 = 0.0486
Number containing exactly two non-zero
digits means two digits are zero and two digits are not zero.
6 possible cases:
0 0 _ _
0 _ 0 _
0 _ _ 0
_ 0 _ 0
_ 0 0 _
_ _ 0 0
Or, recognise that there are two identical zeroes. We need to arrange these zeroes.
So number of ways to arrange
= 4 C 2 = 6
For each case, there are 9 ways to choose each remaining digit (1-9 only, has to be two non-zero digits)
So 9 ways for one digit x 9 ways for the other digit x 6 cases = 486.
Provability = 486 ÷ 10000 = 0.0486
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Part a
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