Probability

30 times because it landed on heads 20 times, but he flipped the coin 50 times. 20+30=50.

40%

For the sake of ease, I'm going to assume that you mean: if a family were to have four children, what is the probability of having two girls and two boys. (Note: if you actually mean a family of four, you would have to start building in the probability of single-parenthood and same-sex adoption since the parents would be included in the total family size.) One could do this problem the long way by writing out all combinations of children and identifying which one meets the criteria given in the problem and which ones do not. All Possible Combos of 4 Children GGGG BGGG GBGG GGBG GGGB BBGG BGBG BGGB GBGB GBBG GGBB GBBB BGBB BBGB BBBG BBBB Each set that matches our criteria are in bold. There are 6 "successful" pairings, out of a total of 16 possible combinations. Thus the probability is 6/16 or .375 or 37.5% that there will be two girls and two boys. That is a long way of doing this problem however and would not be as quick if we had asked about 10 children instead of only 4. Instead it is easier to think of this problem as the probability of one possible combination being correct then multiplying it by all the possible ways of writing it. You can take any possible combination of B and G as long as there are 2 Bs and 2 Gs. For this problem lets use: BBGG. The probability of this exact combination occurring is 1/16: (1/2) * (1/2) * (1/2) * (1/2) = 1/16 Remember that the probability of a certain series of independent events is equal to the each probability multiplied together. Now we must figure out all the possible ways of writing BBGG. In this case there are 6. So now 6 * (1/6) = 6/16 or 3/8 or .375 or 37.5%

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

Rainfall of a day is cumulative rainfall recorded at fix time for past 24 hours. For example rainfall of 26th September at given place in India is amount of rainfall recorded between 3 GMT of 25-9-09 and 3 GMT of 26-09-09.

However rainfall can commence at any time of day say 5 GMT and continued for 5 GMT of next day. Thus rainfall record gets divided one between 5 GMT to 3 GMT and second 3 GMT of next day to 5 GMT conclusion of rain.

It is obvious that 24 hr rain ( 5 GMT to 5 GMT) is more than that of 1 day rain (either 5 GMT to 3 GMT of next day or 3 GMT of next day to 5 GMT conclusion of rains). One day rain is available with Ordinary raingauges (ORG) while 24 hr rain requires Self Recording Raingauges (SRRG). For Design Storm computations 24 hr rain has significance but data record is less than 1 day record.

As a compromize clock hour correction ( 24 hr rainfall / 1 day rainfall ) is worked out for available record where both ORG and SRRG observations are available. The average clock hour correction is then applied over 1 day PMP to get 24 hr PMP value.

N.Y.Apte

Hydrometeorologist

India meteorological Department

New Delhi

1/13

4 out of 12 chances if each dice had 6 sides. 2 out of 6 chances to simplify it.

hypergeom. f(1;13,3,1) * f(1;12,5,1)

Three results are desirable and there are nine possible ones. That gives 3/9, aka 1/3, aka 33%

The total number of outcomes you could get by flipping a coin 4 times is 2^4 or 16 ways as each coin toss yields two possible outcomes (Heads or Tails) and there are four trials. With that said, you need to find out how many ways there are to get 3 heads or 4 heads. You could use combinations to find this:

n C r = n! / [ r! (n-r)! ]

4 C 3 = 4! / [ 3! (4-3)! ]

4 C 3 = 4! / 3!

4 C 3 = 4 4 C 4 = 4! / [ 4! (4-4)! ]

4 C 4 = 4! / 4!

4 C 4 = 1 The total numbers of ways that one could get at least 3 H heads is equal to five (4 ways to get three heads, and one way to get 4 heads). So the probability of getting at least 3 heads in four tosses is equal to 5/16 or 31.25%.

Odds against A = Probabillity against A / Probability for A Odds against A = (1 - Probabillity for A) / Probability for A 9.8 = (1 - Probabillity for A) / Probability for A 9.8 * Probability for A = 1 - Probability for A 10.8 * Probability for A = 1 Probability for A = 1 / 10.8 Probability for A = 0.0926

probability helps you find the chance of something like when you're rolling a dice you might want to know what is the chance you get a 1 or anyother number or if you want to find out how much of school is already over. I find probability very helpful for quickly changing fractions into deciamls.

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.)