P ( B / A ) = P (A ∩ B) / P (A) |
P (A ∩ B) = P (A) * P ( B / A ) |
Q. Two unbiased coins are tossed find probability of getting one head ?
A. Sample space = { HH , TT , HT , TH }
P ( atmost one head ) = 3/4 - include case zero heads and 1 heads
P ( atleast one tail) = 3/4 - include case 1 tails and 2 tails
P ( no tails) = 1 / 4
Q. Find the probability of getting sum of faces more than 6 when 2 dies are rolled?
A. Sample space = { (1,1)
(1,2) (1,3)
(1,4) (1,5)
(1,6)
(2,1) (2,2) (2,3)
(2,4) (2,5)
(2,6)
(3,1) (3,2) (3,3)
(3,4) (3,5)
(3,6)
(4,1) (4,2) (4,3)
(4,4) (4,5)
(4,6)
(5,1) (5,2) (5,3)
(5,4) (5,5)
(5,6)
(6,1) (6,2) (6,3)
(6,4) (6,5)
(6,6) }
P ( sum more than 6 ) = 21 / 36
P ( sum is multiple of 3 ) = 11 / 36
Q. A bag contains 6 white balls and 4 black balls. Probability that they are of same color.
A. P (event) =
= P (choosing 2 white balls from 6) + P( choosing 2 black from 4)
= (6C2/10C2) + (4C2 / 10C2)
Q. Probability that 2 cards drawn at random from pack of cards are both black, both red , both face cards [queen, king, jack], one face card and other normal, both are aces.
A. n ( sample space ) = sample space size is 52.
P ( both black cards ) = number (black cards) / number (cards) = 26C2/52C2 = 1/2
P ( both red cards ) = 1/2
P ( both face cards) = 12 / 52
P ( one face card and other normal ) = (12C1 * 40C1 ) / 52C2.
P ( both aces ) = 4C2 / 52C2.
Q.There are ten 50 paise coins placed on a table. Six of these show tails, four show heads. A coin is chosen at random and flipped over (not tossed). This operation is performed seven times. One of the coins is then covered. Of the remaining nine coins, five show tails and four show heads. The covered coin shows
Ans.a
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