Find the number of cases in which none of the digits show a '6'.
i.e. all three dice show a number other than '6', 5×5×5=125 cases.
Total possible outcomes when three dice are thrown = 216.
The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.
=216?125=91
The required probability = 91/256
The probability of an impossible event is 0.
The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.
The probability of a certain event is 1.
The possible outcomes are as follows :
5H, 5T, (H, 4T), (T, 4H), (2H, 3T) (3H, 2T), i.e. 6 outcomes in all.
Therefore the probability that head appears an odd number of times = 3/6 =1/2 (In only three outcomes out of the six outcomes, head appears an odd number of times).
Given number of balls = 3 + 5 + 7 = 15
One ball is drawn randomly = 15C1
probability that it is either pink or red =
P( only one of them will be selected) = p[(E and not F) or (F and not E)]
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Total number of elementary events =
Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is .
So,required probability = / = 1/12.
The events of selection of two person is redefined as first is a girl and second is a boy or first is boy and second is a girl or first is a girl and second is a girl.
So the required probability:
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Given total number of smileys = 5 + 6 + 3 = 14
Now, required probability =
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