Question

Two dice are thrown once. Find the probability of obtaining :I) a total of 6 ii) a total of 10 iii) the same number on both dice iv) a total more than 9

Answer 1

Answer:

I hope help you

Step-by-step explanation:

1) 5/36 3)6/36

Answer 2

Answer:

When you throw two dice, there will be 36 combinations, which 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, in total 36 combinations.

These combinations are in way

(number on first dice, number on second dice)

Probability = Number of observations/ Total observations.

So probability of finding total of 6 = P(total of 6) = P1 = ?

Now total of 6 will be possible in cases :

(1,5) (5,1) (2,4) (4,2) (3,3). So there are 5 cases or 5 observations,

P1 = 5/36

So probability of finding total of 10 = P(total of 10) = P2 = ?

Now total of 10 will be possible in cases :

(4,6) (5,5) (6,4). So there are 3 cases or 3 observations,

P2 = 3/36.

So probability of finding same number on both dice = P(same number on both dice) = P3 = ?

Now same number on both sides will be possible in cases :

(1,1) (2,2) (3,3) (4,4) (5,5) (6,6). So there are 6 cases or 6 observations,

P3 = 6/36.

So probability of finding total more than 9= P(total more than 9) = P4 = ?

Now total more than 9 will be possible in cases :

(4,6) (5,5) (5,6) (6,4) (6,5) (6,6). So there are 6 cases or 6 observations,

P4 = 6/36.