Question

a die is thrown twice
1. find the probability of not getting 6 either time
2. 4 will come up at least once

Answer 1

Hey mate .
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Given,

A die is thrown once.

We know that a die has six faces.

And also it is thrown twice.

So,

No. of all possible outcomes are as follows:--

(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, Total outcomes = 6 × 6 = 36

(1) Let, E' be the event of not getting 6 either time ,

i.e. No of outcomes favourable to E' = 25

( See from the table given above )

Now,

The probability of not getting 6 either time , p(E')

= $\frac{25}{36}$

(2) Let , E'' be the event that 4 will come up at least once ,

i.e. No of outcomes favourable to E'' = 11

( See from the table given above )

Thus,

Probability of getting 4 either time P(E") = $\frac{11}{36}$


#racks

Answer 2

Given that number of dice thrown = 2.

Then the number of possible outcomes n(S) = 6^2

                                                                           = 36.


(1) Let A be the favorable outcomes that 6 will not get either time.

Total outcomes where 6 gets either time

= {6,1},{6,2},{6,3},{6,4},{6,5},{6,6},{1,6},{2,6},{3,6},{4,6},{5,6}

= 11.

Then the total outcomes where 6 will not get either time

n(A) = 36 - 11

n(A) = 25.


Therefore P(A) = n(A)/n(S)

                          = 25/36




(2) Let B be the favorable outcomes that 4 will come up at least once:

n(B) = 11 {1,4},{2,4},{3,4},{4,4},{5,4},{6,4},{4,1},{4,2},{4,3},{4,5},{4,6}


Therefore the probability P(B) = n(B)/n(S)

                                                  = 11/36.



Hope this helps!