Question

Two fair dice, one blue and one red, are tossed, and the up face on each die is recorded. Define the following events:

E: {The difference of the numbers is 3 or more}
F: {The numbers are equal}

Find the following probabilities:
(a) P(E)= _____
(b) P(F)= _____
(c) P(EF) = _____

Answer 1

Answer:

Step-by-step explanation:

Given that two fair dice, one blue and one red, are tossed, and the up face on each die is recorded.

a) P(E) = P(the difference of the numbers is 3 or more}

Favourable events are (1,4) (1,5)(1,6) (2,5) (2,6) (3,6) (4,1) (5,1) (5,2) (6,1) (6,2)(6,3)

P(E) = $\frac{12}{36} =\frac{1}{3}$

b)P(F)

Favourable events for F = (1,1) (2,2)...(6,6)

P(F) = $\frac{6}{36} =\frac{1}{6}$

c) P(EF)

There is no common element between E and F

P(EF) =0

Answer 2

Answer:

(a)  $\frac{1}{3}$

(b)   $\frac{1}{6}$

(c)  0

Step-by-step explanation:

After tossing two dice, the possible events may be as per following table.

                                              Die - 1

                   1               2              3          4            5              6

Die 2      

 1               1,1             1,2            1,3         1,4         1,5            1,6

 2              2,1            2,2           2.3        2,4         2,5         2,6

 3              3,1            3,2           3,3        3,4          3,5         3,6

 4              4,1            4,2           4,3        4,4          4,5         4,6

 5              5,1            5,2           5,3        5,4          5,5        5,6

 6              6,1            6,2           6,3        6,4          6,5        6,6

E : (1,4), (1,5), (1,6), (2,5), (2,6), (3,6), (4,1), (5,1) (5,2), (6,1), (6,2), (6,3)

(a) P(E) = $\frac{\text{favorable events}}{\text{total events}}$

    = $\frac{12}{36}$

    =  $\frac{1}{3}$

F : (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)

(b) P(F) =  $\frac{6}{36}$

            =  $\frac{1}{6}$

(c) P(EF) = 0

Even a single event is not common in E and F. Therefore, P(EF) would be 0.