Question

13. You roll a single die numbered from 1 to 6. Is
it likely you will roll a number greater than 4
the first time and a number less than 2 the 2nd?​

Answer 1

Answer:

Step-by-step explanation:

Yes we can get numbers more than 4 in a dice

5 and 6

Yes we can get less the 2 in a dice

1

Answer 2

No , It is unlikely that a number greater than 4the first time and a number less than 2 the 2nd will occur

Given :

A single die numbered from 1 to 6

To find :

Is it likely you will roll a number greater than 4 the first time and a number less than 2 the 2nd

Solution :

Step 1 of 3 :

Find total number of possible outcomes

Here it is given that single die numbered from 1 to 6

It is rolled twice

So the event points are

= { (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)}

So the total number of possible outcomes = 36

Step 2 of 3 :

Find the total number of possible outcomes for the event

Let A be the event that a number greater than 4

the first time and a number less than 2 the 2nd will occur

Then the event points for the event A are (5,1) , (6,1)

So the total number of event points for the event A is 2

Step 3 of 3 :

Find it is likely or unlikely to occur

The probability of the event

$\sf{ =P(A) }$

$\displaystyle \sf{ = \frac{Number \: of \: favourable \: cases \: to \: the \: event \: A }{Total \: number \: of \: possible \: outcomes }}$

$\displaystyle \sf{ = \frac{2}{36} }$

$\displaystyle \sf{ = \frac{1}{18} }$

We know that ,

An unlikely event have a probability between 0 and ½

A likely event have a probability between ½ and 1

$\displaystyle \sf Now \: , \: \: \: 0 < \frac{1}{18} < \frac{1}{2}$

Since probability of the event lies between 0 and ½

Hence we can conclude that ,

No , It is unlikely that a number greater than 4

the first time and a number less than 2 the 2nd will occur

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