Solved: Probability of Achieving 6 in 24 Throws of 2 Dice

In summary, the conversation discusses the probability of achieving at least one pair of sixes when rolling two dice 24 times. The attempt at a solution involves calculating the probability of getting a pair of sixes in one throw and the probability of not getting a pair of sixes in 24 throws. However, the problem actually asks for the probability of getting at least one pair of sixes in 24 throws, so this approach is not applicable.
  • #1
sacuan
2
0

Homework Statement



probability to achieve a couple of 6 launching a couple of dies for 24 times

Homework Equations



why i cannot calculate this as:
1/36 * (35/36)EXP23 * (24!/23!)


The Attempt at a Solution



it should be correct:
a="to achive a couple of 6 in 1 throw"
e="never achive a coule of 6 in 24 throws"
i can calculare p(a)=1/36
i can calculate p(e)=[1-p(a)]EXP24
so 1-p(e) should be my correct result
 
Physics news on Phys.org
  • #2
Unless I'm missing something, it seems you can treat this like a Bernoulli scheme with the probability p = 1/36.
 
  • #3
i understand by myself

the exact text is of the problem is:

probability to achieve AT LEAST a couple of 6 launching a couple of dies for 24 times

so now i understand why i cannot use it

thank you
 

Related to Solved: Probability of Achieving 6 in 24 Throws of 2 Dice

1. What is the probability of achieving 6 in 24 throws of 2 dice?

The probability of achieving 6 in 24 throws of 2 dice is approximately 0.315 or 31.5%. This means that in a large number of trials, we can expect to see a 6 appear in about 31.5% of them.

2. How did you calculate this probability?

To calculate the probability, we first need to determine the total number of possible outcomes when throwing 2 dice, which is 36 (6 possible outcomes for the first die, multiplied by 6 possible outcomes for the second die). Out of these 36 outcomes, there are 11 ways to achieve a sum of 6 (1+5, 5+1, 2+4, 4+2, 3+3, 1+5, 5+1, 2+4, 4+2, 3+3, 6+0, 0+6). Therefore, the probability is 11/36, which simplifies to approximately 0.315 or 31.5%.

3. Is this probability affected by the order in which the dice are thrown?

No, the probability remains the same regardless of the order in which the dice are thrown. In other words, whether the first die shows a 1 and the second die shows a 5, or vice versa, the outcome is still considered a sum of 6 and it is included in the total number of possible outcomes.

4. What other factors could affect the probability of achieving 6 in 24 throws of 2 dice?

The probability could be affected by the type and quality of the dice, the surface on which they are thrown, and the skill or technique of the person throwing them. Additionally, if only a limited number of trials are conducted (less than 24 throws), the actual probability may differ from the expected value.

5. Can this probability be applied to other sums or numbers?

Yes, this method of calculating probability can be applied to any sum or number that can be achieved by throwing 2 dice. The only difference would be the total number of possible outcomes and the number of ways to achieve the specific sum or number. For example, the probability of achieving a sum of 7 in 24 throws of 2 dice would be 6/36 or approximately 0.167 or 16.7%.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
2K
  • Precalculus Mathematics Homework Help
Replies
1
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
11
Views
2K
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
5K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top