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you two were be able to give me the value of the denominator before the exam is finish,

but you didn't help me,

are you always answer the questions in this way ?

you are too late, my answer in the exam was p = 2 / (36-4)

**Hannibal lecter**- Replies: 1

Five balls (2 red , 1 blue , 1 green , 1 yellow )

are arranged in a circle, Find the probability that the red balls are not together

please help me with the answer, my exam is about 11 hours

is n = 3 or r = 3 ? if the coin tossed 3 times , or it's not either

are you there bob bundy, can you tell what is the denominator value please is it 36 or not

I'll see that after few hours, but

do the formula you provide mean like that following picture,

( please look at it, is is the same as you wrote)

nCr x (1/2)power r x (1/2) power(n-r)

can you write down this rule again more clearly please,

I can't not understand what is x (1/2)power r please.. my exam is tomorrow

is x a variable or multiplication symbol,

please I won't ask too much again, but the formula is not clear can you type it clearly or like image or anything...

and what will be n value and the k value ?

then what is the denominator ?

and please give me the full answer for all the questions a b and c, because I have an exam please! just this time

upload photo album

yes but He is asking for a general rule!

is there a general rule for that?

there is 2,

do you mean the solution is 2/36??????????? really just that?

**Hannibal lecter**- Replies: 10

when you toss a coin three times, Let x denote the number of heads that appear

find p(x)?

!!!!

is there a general law for this question? how to find this probability?

it's really harddddddddd

**Hannibal lecter**- Replies: 9

Hi, two dice are thrown, fine the probability that the sum is 8 if you know one of them is 5

How did you calculate this combinations by its rule?

or you just know everything about the dice

please tell me how did you calculate this using the rule of combinations (n!/r!(n − r)!)

2 (P= 1/6) the second a 4 (P=1/6) and the third a 1 (P=1/6), is the number gender is the 6 possibilities or the 6 digit of the die?

we didn't use the combinations , you just solve the question before calculating the combinations,

and now if we allowed to re-arrange them and we know that there is six possibilities what after?

how to start calculating

**Hannibal lecter**- Replies: 1

Hi, how many permutation the set has X= ={1,1,2,2,6}

int repetitions and without repetitions

from where did you get the number 4 / 14 ???

and why didn't you use Combinations Or Permutations rules?

I'm so confused I can't imagine this

how could you avoid repetitions by just divide by 2 or 6!!! isn't there something called combinations without repetition or Permutations without repetition

I don't have a time they need the explanation please help me

I can't imagine how could it involve too many calculation? can you give me a hint

**Hannibal lecter**- Replies: 4

Hi, why I can't calculate the probability of birthday problem without taking the complement ?

for example why I can't calculate the probability of at least 2 shared birthday in a room of 30 people

without taking the complement ?

is it too hard? but how to do it?

can you give me another easy example please, I didn't understand that one, maybe an example of marbles or dice or coins, ..

please just give me an easy example,

and what did you mean when you said 5 of 11 ? is it mean p(n=5,k=11) or vice versa?

and did you use the rule of Permutations with Repetition or without ?

as well as ..

the rule of probability is = Number of ways it can happen (event) /Total number of outcomes( sample space)

is the (Combinations and Permutations) the same as =Total number of outcomes( sample space)? or what..

please help me with that one

is it the same of any other problems?

**Hannibal lecter**- Replies: 7

Hi, what is the relationship between (Combinations and Permutations) AND ( Probability) ????

can anyone help me with that please!!!

**Hannibal lecter**- Replies: 2

Hi, how can the sample space of rolling a two dice 36 ? isn't it 42?

it write the there is a possible to get 1,1 and 2,2 and 3,3 and 4,4 and 5,5 and 6,6

but it's for the (first die,second die),(first die,second die),(first die,second die),(first die,second die),(first die,second die),(first die,second die)

but what if it can be (second die,first die),(second die,first die),(second die,first die),(second die,first die),(second die,first die),(second die,first die)

for example (5,5) for the second die at first and first die the last,

so there is more 6 to calculate not just 36!

36+ 6 = 42

am I right or not please give me an explanation