Question

Raj wakes up in the morning and notices that his digital clock reads 07 2(A)5(B).After noon, he looks at the clock again.what is the probality that
I) The number in column A is 4?
ii) The number in column B is an 8?
iii) The number in column A is less than 6?
iv) The number in column B is greater than 5?​

Answer 1

Answer:

i.  $P(A) = \frac{1}{10}$

ii.  $P(B) = \frac{1}{10}$

iii.   $P(A) = \frac{3}{5}$

iv.   $P(B) = \frac{2}{5}$

Step-by-step explanation:

Given

Time: 07 2(A) 5(B)

Calculating (a)

First, we need to list out the possible sample space, S of A

$S = \{0,1,2,3,....,9\}$

$n(S) = 10$

Probability of A being a 4 is the number of occurrence of 4 divided by the number of sample space

$A = \{4\}$

$n(A) = 1$

Hence;

$P(A) = \frac{n(A)}{n(S)}$

$P(A) = \frac{1}{10}$

Calculating (b)

First, we need to list out the possible sample space, S of B

$S = \{0,1,2,3,....,9\}$

$n(S) = 10$

Probability of B being a 8 is the number of occurrence of 8 divided by the number of sample space

$B = \{8\}$

$n(B) = 1$

Hence;

$P(B) = \frac{n(B)}{n(S)}$

$P(B) = \frac{1}{10}$

Calculating (c)

Using the sample space in (a)

$n(S) = 10$

Probability of A being less than 6 is the number of occurrence of less than 6 divided by the number of sample space

$A = \{0,1,2,3,4,5\}$

$n(A) = 6$

Hence;

$P(A) = \frac{n(A)}{n(S)}$

$P(A) = \frac{6}{10}$

$P(A) = \frac{3}{5}$

Calculating (d)

Using the sample space in (b)

$n(S) = 10$

Probability of B being greater than 5 is the number of occurrence of greater than 5 divided by the number of sample space

$B = \{6,7,8,9\}$

$n(B) = 4$

Hence;

$P(B) = \frac{n(B)}{n(S)}$

$P(B) = \frac{4}{10}$

$P(B) = \frac{2}{5}$