Three dice are thrown simultaneously what is the probability of obtaining a total of 17 or 18

Probability for rolling three dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each (three) dies.

When three dice are thrown simultaneously/randomly, thus number of event can be 63 = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its faces.

Worked-out problems involving probability for rolling three dice:

1. 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.

(v) getting a total of atmost 6.

(vi) getting a total of at least 6.

Solution:

Three different dice are thrown at the same time.

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

(i) getting a total of 5:

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

Number of favorable outcomes
P(E1) = Total number of possible outcome = 6/216 = 1/36

(ii) getting a total of atmost 5:

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

Number of favorable outcomes
P(E2) = Total number of possible outcome = 10/216 = 5/108

(iii) getting a total of at least 5:

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

Number of favorable outcomes
P(E3) = 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) getting a total of 6:

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

Number of favorable outcomes
P(E4) = Total number of possible outcome = 10/216 = 5/108

(v) getting a total of atmost 6:

Number of events of getting a total of atmost 6 = 20

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), (1, 2, 2), (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 atmost 6

Number of favorable outcomes
P(E5) = Total number of possible outcome = 20/216 = 5/54

(vi) getting a total of at least 6:

Number of events of getting a total of less than 6 (event of getting a total of 3, 4 or 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), (1, 2, 2), (2, 1, 2), (2, 2, 1).

Therefore, probability of getting a total of less than 6

Number of favorable outcomes
P(E6) = Total number of possible outcome = 10/216 = 5/108

Therefore, probability of getting a total of at least 6 = 1 - P(getting a total of less than 6)

= 1 - 5/108

= (108 - 5)/108

= 103/108

These examples will help us to solve different types of problems based on probability for rolling three dice.

Probability

Random Experiments

Experimental Probability

Events in Probability

Empirical Probability

Coin Toss Probability

Probability of Tossing Two Coins

Probability of Tossing Three Coins

Complimentary Events

Mutually Exclusive Events

Mutually Non-Exclusive Events

Conditional Probability

Theoretical Probability

Odds and Probability

Playing Cards Probability

Probability and Playing Cards

Probability for Rolling Two Dice

Solved Probability Problems

Probability for Rolling Three Dice

9th Grade Math

From Probability for Rolling Three Dice to HOME PAGE

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

Share this page: What’s this?

If three dices are thrown simultaneously, then all the possible outcomes = 63 = 216
∴ Total number of possible outcome, n(S) = 216Let A = event of getting a sum of numbers on three dice as 17 or 18Then the favourable outcomes are given asA = {(6, 6, 5), (6, 5, 6), (5, 6, 6), (6, 6, 6)}

Number of favourable outcomes, n(A) = 4

Hence, required probability, P(A) = P (sum of the numbers on three dices as 17 or 18) = \[\frac{4}{216} = \frac{1}{54}\]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!