If a coin is tossed for 30 times what is the probability of having no heads?
No heads means that every toss lands tails. (0.5)30= 9.3 x 10-10 Note that 109 = 1 trillion, so the probability can be stated this event is likely to occur about 9 times in 10 trillion tosses.
The probability of getting all 10 questions right is, P(10) =
(1/2)10 =
0.0009765...
9 questions right, P(9) =
10C9 ∙(1/2)10 =
10∙(1/2)10 ~ 0.009765...
8 questions right, P(8) =
10C8 ∙(1/2)10 ~ 45∙(1/2)10 ~ 0.043945...
7 questions right, P(7) =
10C7 ∙(1/2)10 ~ 120∙(1/2)10 ~ 0.1171875...
The probability of passing the test with any of the grade is the sum of all the
above; P(passing)
~ 0.1719 ~ 17.2%
What does The first law of probability state?
States that the results of one chance event have no effect on the results of subsequent chance events.
If the odds against an event are 9.8 find the probability that the event will occur?
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.
What is the probability of tossing a fair coin ten times and getting ten heads in a row?
Answer this Question : Probability of getting 10 heads in a row is
(1/2)^10 = 1/1024 = 0.000976 or 0.098 %What is the probability number is perfect square from 10 to 20?
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).
An ordinary six-sided dice is thrown what is the probability that the score is greater than 3?
P(score>3)=1/6+1/6+1/6=3/6=1/2
An ordinary six-sided dice is thrown what is the probability that the score is at least 5?
one in six
An ordinary six-sided dice is thrown what is the probability that the score is less than 3?
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
You are dealt one card from a 52 deck of cards what is the probability of being an ace?
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
How many red cards are in a deck of cards?
In a standard deck of playing cards, there are 26 red cards.
The actual result of this scenario would change depending on how much you simplify it. If you were to simply regard each key as having an equal probability of being struck, then you could say that each key has a 1/88 chance of being hit. That is, statistically speaking, if you were to hit the piano randomly 88 times you should hit each key once. However, this is probably not true in real life. From the vantage point you have over the piano, you are situated more comfortably over the middle range of keys. So this means that you will be more likely to hit those within your range of motion. Another facet to this is the predictability of human nature. Knowing that they will be expected to hit the keys that lie closer to them, the general person will deliberately reach out and hit a key further away from them to 'make it look random'. Contrary to their intent, this actually alters the probability of hitting the other keys, biasing the experiment.
What is the probability of choosing a ten or a jack from a standard deck of playing cards?
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%.
What is the formula to find the probability?
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.)
33.333%.
This can be calculated in two different ways:
1) The likelihood of the first coin being a state quarter is 2 out of 3, or 66.666%. The likelihood of the second coin being a state quarter is 1 out of 2 (since there are now only two coins to pick from, one a state quarter and the other not), or 50.000%. The likelihood of them both being state quarters is then 66.666% x 50.000%, which equals 33.333%.
2) This is the same question as "what is the likelihood that the coin not chosen is not a state quarter?", which is 1 out of 3, or 33.333%.