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    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 3Q. Three identical fair dice are rolled once. The probability that the same number will appear on each of them isQ. 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

    View More

    स्रोत : byjus.com

    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

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    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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    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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    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)

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    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:

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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.

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    स्रोत : www.quora.com

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