Given total number of smileys = 5 + 6 + 3 = 14
Now, required probability =
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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=
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.
P( only one of them will be selected) = p[(E and not F) or (F and not E)]
=
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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
Here Sample space S = { HH, HT, T1, T2, T3, T4, T5, T6 }
Let A be the event that the die shows a number greater than 4 and B be the event that the first throw of the coin results in a tail then,
A = { T5, T6 }
B = { T1, T2, T3, T4, T5, T6 }
Therefore, Required probability =
Total number of chairs = (3 + 5 + 4) = 12.
Let S be the sample space.
Then, n(s)= Number of ways of picking 2 chairs out of 12
= 12×11/2×1 = 66
Let n(E) = number of events of selecting 2 chairs for selecting no white chairs.
=> 8C2 = 8×7/2×1 = 28
Therefore required probability = 28/66 = 14/33.
Required probability:
In a circle of n different persons, the total number of arrangements possible = (n - 1)!
Total number of arrangements = n(S) = (15 ? 1)! = 14 !
Taking three persons as a unit, total persons = 13 (in 4 units)
Therefore no. of ways for these 13 persons to around the circular table = (13 - 1)! = 12!
In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit =
n(E) = 12! X 3!
Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 12! X 3 ! / 14!
= 12!x3x2 / 14x13x12! = 6/14x13 = 3/91
The 8 letters can be written in 8! ways.
n(S) = 8!
Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.
Now the letters a, (bcde), f, g and h can be arranged in 5! ways.
The letters b,c,d and e can be arranged themselves in 4! ways.
n(E) = 5! x 4!
Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14
Hence the answer is 1/14.
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