Box-I contains 8 red, 3 blue and 5 green balls,

Box-II contains 24 red, 9 blue and 15 green balls,

Box-III contains 1 blue, 12 green and 3 yellow balls,

Box-IV contains 10 green, 16 orange and 6 white balls.

A ball is chosen randomly from Box-I; call this ball $b$. If $b$ is red then a ball is chosen randomly from Box-II, if $b$ is blue then a ball is chosen randomly from Box-III, and if $b$ is green then a ball is chosen randomly from Box-IV. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened, is equal to

Two players, $$P_{1}$$ and $$P_{2}$$, play a game against each other. In every round of the game, each player rolls a fair die once, where the six faces of the die have six distinct numbers. Let $$x$$ and $$y$$ denote the readings on the die rolled by $$P_{1}$$ and $$P_{2}$$, respectively. If $$x>y$$, then $$P_{1}$$ scores 5 points and $$P_{2}$$ scores 0 point. If $$x=y$$, then each player scores 2 points. If $$x < y$$, then $$P_{1}$$ scores 0 point and $$P_{2}$$ scores 5 points. Let $$X_{i}$$ and $$Y_{i}$$ be the total scores of $$P_{1}$$ and $$P_{2}$$, respectively, after playing the $$i^{\text {th }}$$ round.

List-I | List-II |
---|---|

(I) Probability of $$\left(X_{2} \geq Y_{2}\right)$$ is | (P) $$\frac{3}{8}$$ |

(II) Probability of $$\left(X_{2}>Y_{2}\right)$$ is | (Q) $$\frac{11}{16}$$ |

(III) Probability of $$\left(X_{3}=Y_{3}\right)$$ is | (R) $$\frac{5}{16}$$ |

(IV) Probability of $$\left(X_{3}>Y_{3}\right)$$ is | (S) $$\frac{355}{864}$$ |

(T) $$\frac{77}{432}$$ |

The correct option is:

_{1}= {1, 2, 3}, F

_{1}= {1, 3, 4} and G

_{1}= {2, 3, 4, 5}. Two elements are chosen at random, without replacement, from the set E

_{1}, and let S

_{1}denote the set of these chosen elements. Let E

_{2}= E

_{1}$$-$$ S

_{1}and F

_{2}= F

_{1}$$\cup$$ S

_{1}. Now two elements are chosen at random, without replacement, from the set F

_{2}and let S

_{2}denote the set of these chosen elements.

Let G

_{2}= G

_{1}$$\cup$$ S

_{2}. Finally, two elements are chosen at random, without replacement, from the set G

_{2}and let S

_{3}denote the set of these chosen elements.

Let E

_{3}= E

_{2}$$\cup$$ S

_{3}. Given that E

_{1}= E

_{3}, let p be the conditional probability of the event S

_{1}= {1, 2}. Then the value of p is

_{1}and C

_{2}be two biased coins such that the probabilities of getting head in a single toss are $${{2 \over 3}}$$ and $${{1 \over 3}}$$, respectively. Suppose $$\alpha $$ is the number of heads that appear when C

_{1}is tossed twice, independently, and suppose $$\beta $$ is the number of heads that appear when C

_{2}is tossed twice, independently. Then the probability that the roots of the quadratic polynomial x

^{2}$$-$$ ax + $$\beta $$ are real and equal, is