The answer will depend on what the trial is!
The probability of getting only one tails is (1/2)7. With seven permutations of which flip is the tails, this gives a probability of: P(six heads in seven flips) = 7*(1/2)7 = 7/128
The probability of getting a diamond and a black seven is zero. Diamonds are red.
Probability of drawing a 7 or 9 is the probability of draw a 7 plus the probability of draw a 9 which is 4/52 + 4/52 = 8/52 = 2/13 = 0.1538.
That's the same as the total probability (1) minus the probability of seven heads. So: 1 - (1/2)7 = 127/128
The probability is 7/36.
four out of seven times
The probability would be once in 128 attempts. You don't have to toss seven coins simultaneously. the 7 tosses just have to be independent of one another.
For me I think it is conceptually easier to think about the probability that the number will contain the digit seven (and the probability that it does not contain the digit 7 is simply one minus the probability that it does). P(number will contain 7) = P(number is in the seven hundreds) + P(number is not in seven hundreds)*[P(number is in the X hundred seventies)+P(number is not in the X hundred seventies)*P(number ends in seven)] So essentially I am considering all of the numbers in the range that start with seven (i.e., are in the seven hundreds), then all of the numbers in the range that aren't in the seven hundreds but have a 7 in the tens place (i.e., the 170s, 270s, etc., and finally all the numbers that don't have a 7 in the hundred or tens place, but that end in 7). Plugging the numbers into my formula above, I get (100/900)+(800/900)*((10/100)+(90/100)(1/10)) = 7/25 is probability that the number does contain a 7, and 1-(7/25)=18/25 is probability that it does not.
The probability of rolling a seven with one roll of a standard number cube is zero.
7*(1/2)7 = 7/128 = 5.47% approx.
So I guess the counters are numbered 1 through 7, and you pick one of them. There are 7 possibilities, 3 of them are even numbers {2,4,6}, so the probability is 3 out of 7. 3/7 = approx. 42.86%
It is 1: a certainty. A complete pack of cards must have a seven.