P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize)
=
Clearly, there are 52 cards, out of which there are 12 face cards.
P (getting a face card) = 12/52=3/13.
Here S= {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}.
Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}.
P(E) = n(E) / n(S)
= 4/8= 1/2
Let number of balls = (6 + 8) = 14.
Number of white balls = 8.
P (drawing a white ball) = 8 /14 = 4/7.
Let S be the sample space.
Then, n(S) = =(52 x 51)/(2 x 1) = 1326.
Let E = event of getting 1 spade and 1 heart.
n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = = 169.
P(E) = n(E)/n(S) = 169/1326 = 13/102.
Let A be the event of getting two oranges and
B be the event of getting two non-defective fruits.
and be the event of getting two non-defective oranges
=
P(first letter is not vowel) =
P(second letter is not vowel) =
So, probability that none of letters would be vowels is =
Possible outcomes = (RY, YP, PR)
2C1 4C1 + 4C1 6C1 + 6C1 2C1
Required probability = (2C1 4C1 + 4C1 6C1 + 6C1 2C1)/(12C2)
= 8 + 24 + 12/66
= 44/66
= 2/3.
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
P(E) =n(E)/n(S)=7/8.
Total number of elementary events =
Given,third ticket =30
=> first and second should come from tickets numbered 1 to 29 = ways and remaining two in ways.
Therfore,favourable number of events =
Hence,required probability = =551 / 15134
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