Answer:
Step-by-step explanation:
Given that two digit numbers are used and one digit is drawn randomly.
Sample space = {10,11,...99}
n(S) = 90
32 is only one number favourable
Hence P(32) = 1/90
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Odd numbers are exactly 45.
Hence prob (odd number) = 45/90 = 1/2
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Multiples of 5 are 10,15,.....95
There are exactly 18 numbers.
Hence P(a multiple of 5) = 18/90 =1/5
The probability of p(32) is 1/90.
The probability of p(odd number ) is 1/2.
The probability of p (a multiple of 5 ) is 1/5.
A random two-digit number (10-99) is drawn.
The probability of;
p (32)
p (odd number )
p (a multiple of 5 )
The sample is (10, 11, 12....., 98. 99)
n(s) = 90
1. The probability of 32 is.
[tex]\rm p(n) = \dfrac{Favourable \ number}{Total \ number}\\\\ p (32) = \dfrac{1}{90}\\ \\ [/tex]
The probability of p(32) is 1/90.
2. The probability of odd numbers.
Total odd numbers = 45
[tex]\rm p(odd \ numbers ) = \dfrac{Odd \ number}{Total \ number}\\\\p(odd \ numbers ) = \dfrac{45}{90}\\\\ p(odd \ numbers ) = \dfrac{1}{2}[/tex]
The probability of p(odd number ) is 1/2.
3. The probability of p (a multiple of 5 ) is,
The multiples of 5 are 5, 10, 15,......90, 95
p (a multiple of 5 ) = 18
[tex]\rm p( (a \ multiple \ of \ 5 ) ) = \dfrac{ (a \ multiple \ of\ 5 )}{Total \ number}\\\\p(odd \ numbers ) = \dfrac{18}{90}\\\\ p(odd \ numbers ) = \dfrac{1}{5}[/tex]
The probability of p (a multiple of 5 ) is 1/5.
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