Question

Two unbiased dice are thrown simultaneously what is the probability of getting atleast one 4 in a single throw of the two dice

Answer 1

Since the probability of 1 die does not affect the probability on the other die (in maths the dice are described as independent), we can just multiply the probabilities of each dice together to get the answer.

To calculate the probabilities of each we can use the formula:

Probability= # of desired outcomes ÷ # of all possible outcomes

What is the probability of getting a multiple of 2? Well we'll list all desired outcomes and count them. The desired outcomes are {2,4,6}, so there are 3 desired outcomes. There are 6 possible outcomes as its a 6 sided die, so the probability is 3/6=1/2

What is the probability of a multiple of 3? The desired outcomes are {3,6} so there are 2 desired outcomes. There are 6 possible outcomes so the probability is 2/6=1/3

Finally to work out the probability of getting a multiple of 2 on the first die and a multiple of 3 on the second we multiply each probability:

Probability= 1/3 × 1/2 = (1×1)/(3×2) = 1/6

We can double check this using a formula we've used before. We can count the desired outcomes and divide by the total outcomes. Here (3,4) means 3 on dice 1 and 4 on dice 2. The desired outcomes are:

{(2,3),(2,6),(4,3),(4,6),(6,3),(6,6)} so there are 6

The possible outcomes are:

{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),…,(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)} so there are 36 (count them all in your head if you don't believe me)

So probability=6/36=1/6 which is what we got earlier so we know we're correct

Answer 2

Explanation:

Two dice are thrown

Here, 2 dice are throw, so possible outcomes are:

In a throw of pair of dice, total no of possible outcomes=36(6×6) 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)

Hence total number number of outcomes =36

Let E= Event of getting the total of numbers on the dice is greater than 10

Here, the total number of number on the dice greater than 10 are (5,6),(6,5) and Hence outcomes favorable to E=3

Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes) x=12