The probability the shooter makes both shots is .7 * .7 = .49, and the probability of making neither is .3 * .3 = .09. So the probability of making exactly 1 out of 2 is 1 - .49 - .9 = .42, or 42 percent.
It is 3/13. The fact that the card is black makes no difference since the probability is the same for both colours.
That probability is the product of the probabilities of the two individual events; for example, if event A has a probability of 50% and event B has a probability of 10%, the probability that both events will happen is 50% x 10% = 5%.
Both are measures of the likelihood of events.
There is a 2/6 probability of that happnening
The probability the shooter makes both shots is .7 * .7 = .49, and the probability of making neither is .3 * .3 = .09. So the probability of making exactly 1 out of 2 is 1 - .49 - .9 = .42, or 42 percent.
Seems like the probability for a single shot is 0.6 So the probability of success with shot 1 is 0.6 and the probability of failing with the first and winning with the second is 0.4x0.6=0.24, same for the other way around, and the probability if winning both times is 0.6x0.6=0.36, and the probability of failing both times is 0.4x0.4=0.16, which is the same as (1-2x.24-.36) Multiply the probability by 100 to get percent.
It is 3/13. The fact that the card is black makes no difference since the probability is the same for both colours.
They are both measures of probability.
That probability is the product of the probabilities of the two individual events; for example, if event A has a probability of 50% and event B has a probability of 10%, the probability that both events will happen is 50% x 10% = 5%.
The probability that he will not win both games is 0.58
Both are measures of the likelihood of events.
Assuming the alternator's failures are unrelated, the probability of both failing is the product of the individual probability, or 0.022, or 0.0004. The duration of the flight does not matter.
If both tosses are fair, the probability of that outcome is one in four.
There is a 2/6 probability of that happnening
they are alike because they both have the last word "probability"
The answer depends on what you mean by "do". Does it mean calculate individually, calculate the probability of either one or the other (or both), calculate the probability of both, calculate some function of both (for example the sum of two dice being rolled)?