Question

Three dice are thrown together. Find the probability of:

(i) getting a total of 5 (ii) getting a total of atmost 5

(iii) getting a total of at least 5. (iv) getting a total of 6.​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

(i) 1/36

(ii) 5/108

(iii) 53/54

(iv) 5/108

Step-by-step explanation:

Three different dice are thrown at the same time.

Therefore, total number of possible outcomes will be 63 = (6 × 6 × 6) = 216.

(i) Number of events of getting a total of 5 = 6

i.e. (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (2, 1, 2) and (1, 2, 2)

Therefore, probability of getting a total of 5

P(E) =   Number of favorable outcomes  / Total number of possible outcome

     = 6/216

     = 1/36

(ii) Number of events of getting a total of atmost 5 = 10

i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1) and (1, 2, 2).

Therefore, probability of getting a total of atmost 5

P(E) =   Number of favorable outcomes  / Total number of possible outcome

= 10/216

     = 5/108

(iii) Number of events of getting a total of less than 5 = 4

i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1) and (2, 1, 1).

Therefore, probability of getting a total of less than 5

P(E) =   Number of favorable outcomes  / Total number of possible outcome

= 4/216

 = 1/54

Therefore, probability of getting a total of at least 5 = 1 - P(getting a total of less than 5)

= 1 - 1/54

= (54 - 1)/54

= 53/54

(iv)

Number of events of getting a total of 6 = 10

i.e. (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2).

Therefore, probability of getting a total of 6

P(E) =   Number of favorable outcomes  / Total number of possible outcome

 = 10/216

     = 5/108