# Two dice, one blue and one grey, are thrown at the same time. Complete the following table:

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Two dice, one blue and one grey, are thrown at the same time. Complete the following table:

 Event: ‘Sumon two dice’ 2 3 4 5 6 7 8 9 10 11 12 Probability

From the above table a student argues that there are 1 1 possible outcomes 2,3,4,5,6,7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability j-j . Do you agree with this argument?

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Total no. of possible outcomes when 2 dice are thrown = 6×6=36 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) }

E ⟶ event of getting sum on 2 dice as 2

No. of favourable outcomes = 1{(1, 1)}

Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)

P(E) = 1/36

E ⟶ event of getting sum as 3

No. of favourable outcomes = 2 {(1, 2) (2, 1)}

P(E) = 2/36

E ⟶ event of getting sum as 4

No. of favourable outcomes = 3 {(3, 1) (2, 2) (1, 3)}

P(E) = 3/36

E ⟶ event of getting sum as 5

No. of favourable outcomes = 4 {(1, 4) (2, 3) (3, 2) (4, 1)}

P(E) = 4/36

E ⟶ event of getting sum as 6

No. of favourable outcomes = 5 {(1, 5) (2, 4) (3, 3) (4, 2) (5, 1)}

P(E) = 5/36

E ⟶ event of getting sum as 7

No. of favourable outcomes = 6 {(1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)}

P(E) = 6/36

E ⟶ event of getting sum as 8

No. of favourable outcomes = 5 {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}

P(E) = 5/36

E ⟶ event of getting sum as 9

No. of favourable outcomes = 4 {(3, 6) (4, 5) (5, 4) (6, 3)}

P(E) = 4/36

E ⟶ event of getting sum as 10

No. of favourable outcomes = 3 {(4, 6) (5, 5) (6, 4)}

P(E) = 3/36

E ⟶ event of getting sum as 11

No. of favourable outcomes = 2 {(5, 6) (6, 5)}

P(E) = 2/36

E ⟶ event of getting sum as 12

No. of favourable outcomes = 1 {(6, 6)}

P(E) = 1/36

 Event ‘Sumon two dice’ 2 3 4 5 6 7 8 9 10 11 12 Probability 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

No, the outcomes are not equally likely from the above table we see that, there is different probability for different outcome.