# two dice are rolled simultaneously. what is the probability that 6 will come up at least once?

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## Two fair dice are rolled simultaneously. The probability that 5 will come up at least once is:

Two fair dice are rolled simultaneously. The probability that 5 will come up at least once is:

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Two fair dice are rolled simultaneously. The probability that 5 will come up at least once is:

Question

Two fair dice are rolled simultaneously. The probability that

5

will come up at least once is:

Open in App Solution

**Probability of getting**

5

**at least once when two dice are rolled simultaneuously.**

The probability of any event

E

is the ratio of the number of favorable outcomes to the total number of sample spaces.

i.e.,

P(E) = Number of favourable outcomesTotal number of outcomes

The number of sides in a die is

6 .

The total outcome for rolling two fair dice simultaneously is

6×6 =36

The possible outcomes for getting at least

5

by rolling a fair dice twice are:

(1,5), (5,1), (2,5), (5,2), (3,5), (5,3), (4,5), (5,4), (5,5), (6,5), (5,6)

So, the number of favorable outcomes is

11 Consider E

to be the probability of getting

5 at least once.

Probability of getting

5 at least once:

P(E)=Number of favourable outcomesTotal number of outcomes

= 1136

**Hence, the probability that 5 will come up at least once is**

1136

**.**

Suggest Corrections 1

SIMILAR QUESTIONS

**Q.**

A die is thrown twice. What is the probability that

(i) 5 will not come up either time?

(ii) 5 will come up at least once?

[Hint: Throwinga die twice and throwing two dice simultaneously are treated as the same experiment].

**Q.**Two dice are thrown simultaneously. What is the probability that:

(i) 5 will not come up on either of them?

(ii) 5 will come up on at least one?

(iii) 5 will come up at both dice?

**Q.**Two fair dice are rolled simultaneously. The probability ofgetting the sum as 3

**Q.**Three identical fair dice are rolled once. The probability that the same number will appear on each of them is

**Q.**Two dice are thrown simultaneously. What is the probability that:

(i) 5

will not come upon either of them

(ii) 5

will come upon at least one

(iii) 5

will come up at both dice

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## Two dice are thrown simultaneously. What is the probability that:(i) 5 will not come upon either of them (ii) 5 will come upon at least one (iii) 5 will come up at both dice

Click here👆to get an answer to your question ✍️ Two dice are thrown simultaneously. What is the probability that:(i) 5 will not come upon either of them (ii) 5 will come upon at least one (iii) 5 will come up at both dice

Two dice are thrown simultaneously. What is the probability that:Question

(i) 5 will not come upon either of them

(ii) 5 will come upon at least one

(iii) 5 will come up at both dice

Medium Open in App Solution Verified by Toppr

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)

**Solution(i):**

Let E be the event of not getting a 5 on either of the two dice

No. of favorable outcomes = 25

(i.e.,(1,1)(1,2)(1,3)(1,4)(1,6)(2,1)(2,2)(2,3)(2,4)(2,6)(3,1)(3,2)

(3,3)(3,4)(3,6)(4,1)(4,2)(4,3)(4,4)(4,6)(6,1)(6,2)(6,3)(6,4)(6,6) )

P(E)= 36 25

**Solution(ii):**

Let E be the event of getting a 5 at least once

No. of favorable outcomes =11(i.e.,(1,5)(2,5)(3,5)(4,5)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,5)

P(E)= 36 11

**Solution(iii):**

Let E be the event of getting a 5 on both dice

No. of favorable outcomes =1(i.e.,(5,5)

P(E)= 36 1

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## Two dice are rolled simultaneously. What is the probability that one of them shows a number which is at least 4 and the other shows a number which is at most 3?

Answer (1 of 2): One dice event A = 1,2,3 atmost 3 2nd dice event B = 4,5,6 atleast 4 P(A) = 3/6 P(B) = 3/6 Required probability = P(A) * P(B) * 2. Multiply by two since A and B dice can be exchanged.

Two dice are rolled simultaneously. What is the probability that one of them shows a number which is at least 4 and the other shows a number which is at most 3?

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Sort Ravi Teja

Lives in Hyderabad, Telangana, India1y

One dice event A = 1,2,3 atmost 3

2nd dice event B = 4,5,6 atleast 4

P(A) = 3/6 P(B) = 3/6

Required probability = P(A) * P(B) * 2.

Multiply by two since A and B dice can be exchanged.

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Koyel Pramanick

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Assuming dice are distinguishable:

Here first if all, number of all possible outcomes is = 6×6 = 36

Now, 1st one results in any one of 4, 5, 6 and 2nd one results in any one of 1, 2, 3. (Here as the dice are distinguishable, we can consider any one of them as 1st one and another as 2nd one)

So, here no of favourable cases = 3×3 =9

So, the required probability = 9/36 = 1/4.

Assuming dice are indistinguishable:

Now, number of all possible outcomes = 21

Now, number of favourable cases are also 9 here as here we have to make unordered pairs (x,y) where x can take values from set {1,2,3} and y can take va

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Lance Berg

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Two dice are rolled, their sum is 8. What is the probability that one die shows number 3?

As with many probabilities involving small numbers, you can get this by simply counting.

I’ll assume you mean fair and normally marked six sided dice.

There are three ways to get a total of eight. The largest single die result is a 6, which can be paired with a 2. The second largest single die result is a 5, which pairs with a 3. The third largest single die result is a 4, which pairs with another 4.

The tricky bit is, there’s only one way to get the 4+4, but there are two ways to get each of the other results: 6+2 and 2+6 for example. This means that instead of 3 results (or 6 results with 2 of

Jonathan Fivelsdal

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When two dice are rolled simultaneously, what is the probability that the sum of the number on the upper faces is at the most 5?

You can represent the outcome of the 2 dice as ordered pairs where the first entry is the result of the first die and the second entry is the result of the second die.

Below are the outcomes that will produce a sum between 2 and 5

Sum of two: (1,1)

Sum of three: (1,2) and (2,1)

Sum of four: (1,3), (2,2) and (3,1)

Sum of five: (1,4), (2,3), (3,2) and (4,1)

Adding up the cases, there is a total of 10 outcomes where the sum of the dice is at most 5. Now your sample space is the following:

S = { (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

Christopher Pellerito

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A man throws 4 dice. What is the probability that same number shows up in at least two faces?

There are 6^4 = 1296 total ways to roll four dice.

4 of a kind: 6P1 = 6 ways.

3 of a kind: 6P2 * 4! / 3! = 120 ways.

2 pair: 6P2 * 4! / 2! / 2! / 2! = 90 ways.

1 pair: 6P3 * 4! / 2! / 2! = 720 ways.

0 pair: 6P4 = 360 ways.

You could add up all the cases of one pair or better and get

(6+120+90+720) / 1296 = 936 / 1296 = 13 / 18

Or, of course, you could just recognize that you need the complement of the zero pair case:

1 - (360 / 1296) = 936 / 1296 = 13 / 18

Edward Williams

M.A. from University of Wisconsin - Madison (Graduated 1968)Author has 2.1K answers and 16.2M answer views1y

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A pair of dice is rolled. What is the probability of getting the sum of at most 5?

There are 36 possible outcomes, ranging from (1,1) to (6,6). Of these, (1,1), (1,2), (2,1), (2,2), (3,1), (1,3), (2,3), (3,2), (4,1), and (1,4) sum to 5 or lower. That’s 10/36 or 5/18.

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Ed Palone

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A die is rolled. What is the probability of getting an even number or a number greater than 4?

We can use the formula for the probability of more than one type of successful outcome (“or”), so that you know how to use it in a problem in which the # of successes is not as easy to count.

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