Probability

You cant win

There are 11 numbers from 10 to 20 if we include 10 and 20. Of those, only 16 is a perfect square. So the probability if 1/11 or .09 repeating.

There are 8 marbles in the bag, and 6 are green, so the chance that the first one you pick is green is 6/8 or .75. Let's call the event where you pick the green marble first, G, for green of course. Now since you picked a marble there are only 7 left. If you picked a green one then the chances of picking a purple one are now 2/7 since there are two purple marbles and seven total marbles. Let's call the event of picking the purple marble F, (I was going to use P but we need that letter for probability. Purple is a fine color so I picked F.) Now we use the conditional probability rule that tells us what is the chance of picking purple given that we already picked green. The symbol P(F|G) means probability of event F given that event G has already happened. P(F|G)= (the probability of picking green and purple)/ (probability of picking green.) We know these from above. G=6/8 and If we pick a green, probability of picking a purple is 2/7 so we multiply these to get probability of picking both and we have 6/8x2/7 or 12/56. So 12/56=(Probability of Picking green and purple)/( probability of picking green). We have 12/56=[P(G and F)]/(6/8) we want P(G and F) so we multiply 12/56x 6/8 and we have 72/448 So the answer is : 72/448 or about .16 (NOTE: this would be a totally different problem if we took out the first marble then put it back. It is important to be sure what is being asked. If you replaced the marble, the problem is much easier. It is simply 6/8 x 2/8 =12/64 or 3/16) Some people have trouble remembering or understanding the conditional probability rule. I will take just a second to explain it in the hopes it will make it easier to use and remember. The multiplication rule says if we have two mutually exclusive events, A and B, the probability of A and B is P(A)xP(B), so if we want event A to occur THEN event B, we have P(A)xP(B|A) which means probability of A multiplied by probability of B given A has already happened. This equal probability of A and B so we have: P(A)xP(B|A)=P(A and B) . Now divide by P(A) and we have: P(B|A)=P(A and B)/P(A). This is the way the rule is usually stated. Note: P(A|B)=P(A and B)/P(B).

P(score>3)=1/6+1/6+1/6=3/6=1/2

one in six

There are six possible outcomes of rolling a six sided die.
However, only two of these (1 and 2) are favorable.
So, the probability of rolling less than three is 2/6 = 1/3.

you have a 1 in 6 chance of rolling a 6.

you have a 1 in 13 chance to draw a 10 (there are 4 tens in a pack of 52: thus 52/4 = 13)

(1/6)*(1/13)*100 = 1.282 % of drawing a ten and rolling a 6 at the same time

Probability = Chance of Success / Total Chances (Chance of Success + Chance of Failure) There are 4 aces in a 52 card deck and 48 cards that are not aces. Probability of being dealt an ace = 4 / (4 + 48) = 4/52 = .0769 or about 7.7 percent

There are 4 tens, 4 jacks, and 52 total cards. Assuming you take a single card out of a full and shuffled deck, the chances are 8/52=2/13=15.38%.

Its not an exact thing like with circle radius, but basically add together all the numbers and make a fraction like if there are 3 yellow balls, 3 green ones and 5 blue ones in a bag, then you add 3+3+5 = 11, and then you figure out that the probability of picking a blue ball out of the bag is 5/11.

Probability is the ratio of all outcomes that you define as being of interest divided by all possible outcomes.

For example, the number of ways to get doubles on 2 6-sided dice is 6 (1,1 2,2, 3,3 4,4 5,5 6,6) but there are 36 different ways for the dice to turn up so the probability of getting doubles on a single roll is 6/36 = 1/6

Likewise there are 6 ways to roll a 7 (1,6 2,5 3,4 4,3 5,2 6,1) so the odds of rolling a seven on a single roll are 6/36 = 1/6

The math gets more involved as you start looking at situations where the odds of getting any one of the particular outcomes are not the same - for example, with loaded dice - or where you are looking at a sequence of events.

25% or 1 out of 4

(You can use a Punnett Square for that and/or other questions like it.)

theoretical

A poker deck has four aces.

The probability of getting three fives in the first three rolls and non-fives

in the

next three rolls is; P(5,5,5,N5,N5,N5) =

1/6 x 1/6 x 1/6 x 5/6 x 5/6 x 5/6

=

0.002679...

The number of different order in which the fives can come out is given by;

6C3 =

6!/[3!∙(6-3)!] =

20

So the probability that in 6 rolls of a fair die exactly three fives (in any order) will

come out is; P(three

fives any order) =

(20)∙(1/6)3∙(5/6)3 =

0.05358... ~ 5.4%

1/16

10 :)

The chance is 3.5/7.

the answer is 5/36 coz (2,6) (3,5) (4,4) (5,3) (6,2) which are the five possible outcomes out of the 36 total no. of outcomes. Enjoy!! Kush

Sum of two dice eqauls 8:

2,6

3,5

4,4

5,3

6,2

= 5 in 36

In a line in 6! = 6*5*4*3*2*1 = 720 ways.

it is written like p( ) and don'T improve k bye 8)