Respuesta :
Answer:
1453/23328
Step-by-step explanation:
Since we are going to have at least three 5's to obtain a number divisible by 125, the number of ways we can combine exactly three 5's is ⁶C₃. The number of ways we can combine the remaining dice without containing a five is 5³, since there are 3 remaining dice. So, for exactly three 5's , we have ⁶C₃ × 5³. Similarly, for exactly four 5's, we have ⁶C₄ × 5² (5² since there are two dice left). Also, for exactly five 5's, we have ⁶C₅ × 5¹ (5¹ since there are one die left). Finally, for exactly six 5's, we have ⁶C₆ × 5⁰(5⁰ since there are no dice left without a five).
So, for at least three 5's, we have ⁶C₃ × 5³ + ⁶C₄ × 5² + ⁶C₅ × 5¹ + ⁶C₆ × 5⁰
= 20 × 125 + 15 × 25 + 6 × 5 + 1 × 1
= 2500 + 375 + 30 + 1
= 2906 ways
The total number of ways of combining the six dice is 6⁶
So, the probability of obtaining a product divisible by 6 is thus P = 2906/6⁶ = 2906/46656
= 1453/23328
Using the binomial distribution, it is found that there is a 0.0536 = 5.36% probability that the product is divisible by 125.
The product will be divisible by 125 in two outcomes:
- 3 of the dices result in 5, as [tex]5^3 = 125[/tex]
- 6 of the dices result in 5, as [tex]5^6 = 5^3 \times 5^3[/tex]
For each dice, there are only two possible outcomes, either the result is 5, or it is not. The results of the dices are independent, hence, the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- Six dices are rolled, hence [tex]n = 6[/tex].
- Each number is equally as likely, hence the probability of a 5 being rolled, that is, a success, is [tex]p = \frac{1}{6} = 0.1667[/tex].
The probability that the product is divisible by 125 is:
[tex]p = P(X = 3) + P(X = 6)[/tex]
Hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{6,3}.(0.1667)^{3}.(0.8333)^{3} = 0.0536[/tex]
[tex]P(X = 6) = C_{6,6}.(0.1667)^{6}.(0.8333)^{0} \approx 0[/tex]
Then:
[tex]p = P(X = 3) + P(X = 6) = 0.0536 + 0 = 0.0536[/tex]
0.0536 = 5.36% probability that the product is divisible by 125.
A similar problem is given at https://brainly.com/question/24863377
