What is $P(EF)$? Is this $P(E\cap F)$? What is $(}\genfrac{}{}{0ex}{}{48}{12,12,12,12}{\textstyle )$?

Lucian Maddox
2022-07-02
Answered

What is $P(EF)$? Is this $P(E\cap F)$? What is $(}\genfrac{}{}{0ex}{}{48}{12,12,12,12}{\textstyle )$?

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In a fuel economy study, each of 3 race cars is tested using 5 different brands of gasoline at 7 test sites located in different regions of the country. If 2 drivers are used in the study, and test runs are made once under each distinct set of conditions, how many test runs are needed?

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What is the probability that 2 or more numbers out of 4 are among the drawn lottery numbers (6 out of 49 without replacement)?

asked 2022-07-14

Simple combinatorics and probability theory related question

5 apples are randomly distributed to 4 boxes. We need to find probability that there are 2 boxes with 2 apples, 1 box with 1 apple and 1 empty box.

I'm getting the correct answer with $\frac{\frac{5!}{2!2!1!0!}\ast 4\ast 3}{{4}^{5}}=0.3515625$ (anyway, the answer is said to be 0.35, but I think it is a matter of rounding).

But I don't understand why there are ${4}^{5}$ elementary events in total. Firstly, I thought It should be $(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\textstyle (}\genfrac{}{}{0ex}{}{4}{5}{\textstyle )}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}})$ - number of combinations with repetitions, but I couldn't get the proper answer.

Isn't approach with $(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\textstyle (}\genfrac{}{}{0ex}{}{4}{5}{\textstyle )}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}})$ elementary events more correct?

5 apples are randomly distributed to 4 boxes. We need to find probability that there are 2 boxes with 2 apples, 1 box with 1 apple and 1 empty box.

I'm getting the correct answer with $\frac{\frac{5!}{2!2!1!0!}\ast 4\ast 3}{{4}^{5}}=0.3515625$ (anyway, the answer is said to be 0.35, but I think it is a matter of rounding).

But I don't understand why there are ${4}^{5}$ elementary events in total. Firstly, I thought It should be $(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\textstyle (}\genfrac{}{}{0ex}{}{4}{5}{\textstyle )}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}})$ - number of combinations with repetitions, but I couldn't get the proper answer.

Isn't approach with $(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\textstyle (}\genfrac{}{}{0ex}{}{4}{5}{\textstyle )}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}})$ elementary events more correct?

asked 2022-09-10

Take 13 integers. Prove that if any 12 of them can be partitioned into two sets of six each with equal sums, then all the integers are the same.

asked 2022-07-05

The problem is as follows: We have box of the following:

- {A} - 40W light bulbs, N(A) = 4

- {B} - 65W light bulbs, N(B) = 5

- {C} - 75W light bulbs, N(C) = 6

We select 2 bulbs at random. Given that at least one of these is rated 75W, what is the probability that they are both rated 75W?

Using a probability tree, I was able figure out the answer by dividing the probability of both bulbs being 75W (6/15)*(5/14) by all the cases that included a 75-W bulb including the case where both were 75 W.

However, how should I do this using combinatorics?

- {A} - 40W light bulbs, N(A) = 4

- {B} - 65W light bulbs, N(B) = 5

- {C} - 75W light bulbs, N(C) = 6

We select 2 bulbs at random. Given that at least one of these is rated 75W, what is the probability that they are both rated 75W?

Using a probability tree, I was able figure out the answer by dividing the probability of both bulbs being 75W (6/15)*(5/14) by all the cases that included a 75-W bulb including the case where both were 75 W.

However, how should I do this using combinatorics?

asked 2022-05-21

With a standard 52 deck (13 types and 4 suites), what is the probability that a poker hand contains cards of five different types? A poker hand is a set of 5 cards.

The obvious answer is A / B where A is the number of poker hands that have 5 different types of cards and B is the total number of poker hands which is C(52,5).

Right now I think A = C(13,5) since there are 13 different types of cards I choose 5 from them but I am unsure if A = 4 * C(13,5) instead, where 4 is the different suites.

The obvious answer is A / B where A is the number of poker hands that have 5 different types of cards and B is the total number of poker hands which is C(52,5).

Right now I think A = C(13,5) since there are 13 different types of cards I choose 5 from them but I am unsure if A = 4 * C(13,5) instead, where 4 is the different suites.