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PASSAGE  1
There are n urns, each of these contain n + 1 balls. The ith urn contains i white balls and (n + 1 – i) red balls. Let u_{i} be the event of selecting ith urn, i = 1, 2, 3 .........., n and w the event of getting a white ball.
Q. If P(u_{i}) ∝ i, where i = 1, 2, 3,......., n, then P(w) =(2006  5M, –2)
P(u_{i}) ∝ i ⇒ P(u_{i})= k_{i}, But ∑ P(u_{i})=1
By total prob. theorem
PASSAGE  1
There are n urns, each of these contain n + 1 balls. The ith urn contains i white balls and (n + 1 – i) red balls. Let u_{i} be the event of selecting ith urn, i = 1, 2, 3 .........., n and w the event of getting a white ball.
Q. If P(u_{i}) = c, (a constant) then P(u_{n}/w) = (2006  5M, –2)
P(u_{i}) = c
Using Baye’s theorem,
PASSAGE  1
There are n urns, each of these contain n + 1 balls. The ith urn contains i white balls and (n + 1 – i) red balls. Let u_{i} be the event of selecting ith urn, i = 1, 2, 3 .........., n and w the event of getting a white ball.
Q. Let P(u_{i}) = , if n is even and E denotes the event of choosing even numbered urn, then the value of P(w / E) is (2006  5M, –2)
(n being even)
PASSAGE  2
A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required. (2009)
Q. The probability that X = 3 equals
P ( X = 3) = (probability of not a six in first chance) × (probability of not a six in second chance) × (probability of a six in third chance)
PASSAGE  2
A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required. (2009)
Q. 5. The probability that X ≥ 3 equals
P( X ≥ 3) = 1 (X< 3) = 1 [P( X = 1) + P(X= 2)]
PASSAGE  2
A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required. (2009)
Q.The conditional probability that X ≥ 6 given X > 3 equals
Let us define the events A≡X≥ 6 and B ≡X>3 so that A∩B ≡ X ≥ 6≡A
Now
and P(B)
∴ P(A/B) =
PASSAGE  3
Let U_{1} and U_{2} be two urns such that U_{1} contains 3 white and 2 red balls, and U_{2 }contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U_{1 }and put into U_{2}.
However, if tail appears then 2 balls are drawn at random from U_{1} and put into U_{2} . Now 1 ball is drawn at random from U_{2}. (2011)
Q. The probability of the drawn ball from U_{2} being white is
P (white) = P (H ∩ white) + P ( T ∩ white) = P (H) P (white/H) + P (T) P (white/T)
PASSAGE  3
Let U_{1} and U_{2} be two urns such that U_{1} contains 3 white and 2 red balls, and U_{2 }contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U_{1 }and put into U_{2}.
However, if tail appears then 2 balls are drawn at random from U_{1} and put into U_{2} . Now 1 ball is drawn at random from U_{2}. (2011)
Q. Given that the drawn ball from U_{2 }is white, the probability that head appeared on the coin is
P(H/white) =
PASSAGE  4
A box B_{1} contains 1 white ball, 3 red balls and 2 black balls. Another box B_{2} contains 2 white balls, 3 red balls and 4 black balls. A third box B_{3} contains 3 white balls, 4 red balls and 5 black balls.
Q. If 1 ball is drawn from each of the boxes B_{1}, B_{2} and B_{3}, the probability that all 3 drawn balls are of the same colour is (JEE Adv. 2013)
Probability that all balls are of same colour = P (all red) + P (all white) + P (all black)
PASSAGE  4
A box B_{1} contains 1 white ball, 3 red balls and 2 black balls. Another box B_{2} contains 2 white balls, 3 red balls and 4 black balls. A third box B_{3} contains 3 white balls, 4 red balls and 5 black balls.
Q. If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B_{2} is
Let E_{1}, E_{2}, E_{3} be the events that bag B_{1}, B_{2} and B_{3 }is selected respectively.
Let E be the event that one white and one red ball is selected.
Then by baye’s theorem,
PASSAGE  5
Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let x_{i} be number on the card drawn from the ith box, i = 1, 2, 3. (JEE Adv. 2014)
Q. The probability that x_{1} + x_{2} + x_{3} is odd, is
x_{1} + x_{2} + x_{3} will be odd If two are even and one is odd or all three are odd.
∴ Required probability = P (EEO) + P(EOE) + P(OEE) + P(OQO)
PASSAGE  5
Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let x_{i} be number on the card drawn from the ith box, i = 1, 2, 3. (JEE Adv. 2014)
Q. 12. The probability that x_{1}, x_{2}, x_{3 }are in an arithmetic progression, is
If x_{1}, x_{2}, x_{3 }are in AP then 2x^{2} = x_{1 }+ x_{3 }
∵ LHS is even, x_{1} & x_{3} can be both even or both odd. x_{1 }and x_{3} both can be even in 1 × 3 = 3 ways x_{1} and x_{3 }both can be odd in 2 × 4 = 8 ways
∴ Total favourable ways = 3 + 8 = 11
Also one number from each box can be drawn in 3 × 5 × 7 ways
∴ Total ways = 105 Hence required probability =
PASSAGE  6
Let n_{1} and n_{2} be the number of red and black balls, respectively, in box I. Let n_{3} and n_{4} be the number of red and black balls, respectively, in box II. (JEE Adv. 2015)
Q. One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is , then the correct option(s) with the possible values of n1, n_{2}, n_{3} and n_{4} is(are)
Let E_{1} ≡ box I is selected
E_{2} ≡ box II is selected
E ≡ ball drawn is red
P(E_{2}/E) =
or
On checking the options we find (a) and (b) are the correct options.
PASSAGE  6
Let n_{1} and n_{2} be the number of red and black balls, respectively, in box I. Let n_{3} and n_{4} be the number of red and black balls, respectively, in box II. (JEE Adv. 2015)
Q. A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is , then the correct option(s) with the possible values of n_{1 }and n_{2} is(are)
E_{1} ≡ Red ball is selected from box I
E_{2} ≡ Black ball is selected from box I
E ≡ Second ball drawn from box I is red
∴ P(E) = P(E_{1}) P(E/E_{1}) + P(E_{2}) P(E/E_{2})
On checking the options, we find (c) and (d) have the correct values.
PASSAGE  7
Football teams T_{1} and T_{2} have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T_{1} winning, drawing and losing a game against T_{2} are and respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game.
Let X and Y denote the total points scored by teams T_{1} and T_{2 }respectively after two games.
Q.P (X > Y) is (JEE Adv. 2016)
(X, Y) = {(6, 0), (4, 1), (3, 3), (2, 2), (4, 4), (0, 6)}
P(X > Y) = P(T_{1} wins 2 games or T_{1} win one game other is a draw)
PASSAGE  7
Football teams T_{1} and T_{2} have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T_{1} winning, drawing and losing a game against T_{2} are and respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game.
Let X and Y denote the total points scored by teams T_{1} and T_{2 }respectively after two games.
Q. P (X = Y) is (JEE Adv. 2016)
(X, Y) = {(6, 0), (4, 1), (3, 3), (2, 2), (4, 4), (0, 6)}
P(X = Y) = P(T_{1} wins 1 game loses other game or both the games draw)
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