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A card is drawn at random from a deck of cards. Find the probability of getting 3 of diamond.
a. 1/52
b. 1/38
c. 3/56
d. 3/38
Ans - a
Since a pack consist 52 cards and among that cards there are 13 diamonds.
Now for same space, A card is drawn out of 52 cards i.e
n(S) = (52,a. = n(S) = 52
Now for event for occurring 3 of diamonds in one drawn out of 13 =
n(E) = 1
Hence probability of occurrence of getting 3 of diamond
P(E) = n(E)/n(S)
= 1/52
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A jar contains 3 red marbels , 7 green marbels and 10 white marbles. If a marble is drawn at random , What is the probability that marble drawn is white ?
a. 2/5
b. 1/2
c. 3/8
d. 10/13
Ans - b
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An urn contains 10 black balls and 5 white balls. 2 balls are drawn from the urn one after other without replacement. What is the probability that both drawn are black ?
a. 2/7
b. 3/7
3 4/7
d. 6/7
Ans - b
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A jar contains 3 red marbels, 7 green marbels and 10 white marbles. If a marble is drawn at random, What is the probability that marble drawn is white?
a. 2/5
b. 1/2
c. 3/8
d. 10/13
Ans – 2
Solution :
Here Red = 3
Green = 7
White = 10
Hence total sample space is (3+7+10)= 20
Out of 20 one ball is drawn n(S) = {c(20,a.} = 20
To find the probability of occurrence of one White marble out of 10 white ball
n(R)={c(10,a.} = 10
Hence P(R) = n(R)/n(S)
= 10/20 = 1/2
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We have six students say A, B, C, D, E, F participating in a quiz contest. Out of six students only two can reach to the final. What is the probability of reaching to the final of each student ?
a. 1/2
b. 2/3
c. 1/3
d. 1/6
Ans – c
Since out of 6, 2 can reach the final. Hence sample space is
n(S) = 6 c2 = 6!/(6-b.!×2! = 15
Here event of occurrence of probability of each student out of six (A B C D E F) = (AB AC AD AE AF) = n(E) = 5
Now P(E) = 5/15 = 1/3
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An bag contains 10 black balls and 5 white balls. 2 balls are drawn from the bag one after other without replacement. What is the probability that both drawn are black ?
a. 2/7
b. 3/7
c. 4/7
d. 6/7
Ans - 2
Solution :
Let E and F denote respective events that first and second ball drawn are black.
We have to find here P(E), P(E/F) and P(E n F )
Now P(E) = P(Black in first drawn) = 10/15
Also given that the first ball is drawn i.e events E has occurred. Now there are 9 black balls and 5 white balls left in the urn. Therefore the probability that the second ball drawn is black, given that the ball first drawn is black nothing but conditional probability of F given that E has occurred already.
Hence P(E/F) = 9/14
Now by the multiplication rule of probability
P(E n F) = P(E) × P(E/F)
= 10/15 × 9/14 = 3/7
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