Chevy plays a game in which she throws a pair of fair cubical dice numbered12345624times Find the probability that she throws at least one double six

EASY

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

If

A

and

B

are events with

P (A \cup B) = \frac{3}{4}, P (A^{'}) = \frac{2}{3}

and

P (A \cap B) = \frac{1}{4}

then

P (B)

is

EASY

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

A

and

B

are events such that

P (A \cup B) = \frac{3}{4}, P (A \cap B) = \frac{1}{4}, P (A) = \frac{2}{3}

, then

P (A \cap B)

is

HARD

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

Consider the following events:

$E_{1} :$ Six fair dice are rolled and at least one die shows six
$E_{2} :$ Twelve fair dice are rolled and at least two dice show six
Let $p_{1}$ be the probability of $E_{1}$ and $p_{2}$ be the probability of $E_{2} .$ Which of the following is true ?

HARD

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

Four persons independently solve a certain problem correctly with probabilities

\frac{1}{2}, \frac{3}{4}, \frac{1}{4}, \frac{1}{8}

. Then the probability that the problem is solved correctly by at least one of them is

EASY

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

Two cards are drawn at random and one-by-one without replacement from a well-shuffled pack of

52

playing cards. Find the probability that one card is red and the other is black.

MEDIUM

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

Three players

A, B

and

C

play a game. The probability that

A, B

and

C

will finish the game are respectively

\frac{1}{2}, \frac{1}{3}

and

\frac{1}{4}

.

The probability that the game is finished is.

MEDIUM

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

If the probability of hitting a target by a shooter, in any shot is

\frac{1}{3},

then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than

\frac{5}{6},

is:

EASY

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

If three dices are thrown then the probability that the sum of the numbers on their uppermost faces to be atleast

5

is

MEDIUM

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

A

and

B

are independent events. The probability that both

A

and

B

occur is

\frac{1}{20}

and the probability that neither of them occurs is

\frac{3}{5}

. The probability of occurrence of

A

is

HARD

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

Let

E, F

and

G

be three events having probabilities

P (E) = \frac{1}{8}, P (F) = \frac{1}{6}

and

P (G) = \frac{1}{4},

and let

P (E \cap F \cap G) = \frac{1}{10}

For any event

H

, if

H^{c}

denotes its complement, then which of the following statements is (are) TRUE ?

HARD

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

Three persons P, Q and R independently try to hit a target. If the probabilities of their hitting the target are

\frac{3}{4}, \frac{1}{2}

and

\frac{5}{8}

respectively, then the probability that the target is hit by P or Q but not by R is:

MEDIUM

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

A determinant of second order is made with the elements 0, 1. What is the probability that the determinant is positive?

HARD

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is:

HARD

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

For three events,

A, B

and

C,

P

(Exactly one of

A

or

B

occurs)

= P

(Exactly one of

B

or

C

occurs)

= P

(Exactly one of

C

or

A

occurs)

= \frac{1}{4}

and

P

(All the three events occur simultaneously)

= \frac{1}{16} .

Then the probability that at least one of the events occurs, is:

HARD

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

If

A

and

B

are two events such that

P (A \cup B) = P (A \cap B)

, then the incorrect statement amongst the following statements is :

EASY

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

Let

A

and

B

be two events such that the probability that exactly one of them occurs is

\frac{2}{5}

and the probability that

A

or

B

occurs is

\frac{1}{2}

, then the probability of both of them occur together is

MEDIUM

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

In a box, there are

20

cards, out of which

10

are labelled as

A

and the remaining

10

are labelled as

B

. Cards are drawn at random, one after the other and with replacement, till a second

A

card is obtained. The probability that the second

A

card appears before the third

B

card is:

EASY

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

If $A$ and $B$ are events such that $P (A) = 0.42, P (B) = 0.48$ and $P (A and B) = 0 .16$ . Then,

$I .$ $P (not A) =0 . 58$

$II .$ $P (not B) = 0.52$

$III .$ $P (A or B) = 0.47$

HARD

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than

99 %

is:

HARD

Mathematics>Statistics and Probability>Conditional Probability and the Rules of Probability>Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model

In a class of

60

students,

40

opted for NCC,

30

opted for NSS and

20

opted for both NCC and NSS. If one of these students is selected at random, then the probability that the student selected has opted neither for NCC nor for NSS is :

Chevy plays a game in which she throws a pair of fair cubical dice numbered $\{1, 2, 3, 4, 5, 6\} 24$ times. Find the probability that she throws at least one double six.

Important Questions on Conditional Probability and the Rules of Probability

Learn and Prepare for any exam you want !

Chevy plays a game in which she throws a pair of fair cubical dice numbered 1, 2, 3, 4, 5, 6 24 times. Find the probability that she throws at least one double six.

Important Questions on Conditional Probability and the Rules of Probability

Learn and Prepare for any exam you want !

Chevy plays a game in which she throws a pair of fair cubical dice numbered $\{1, 2, 3, 4, 5, 6\} 24$ times. Find the probability that she throws at least one double six.