Probability — Page 14 | JEE Mathematics

Q131

Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability, that first drawn marble is red and second drawn marble is white, is

A

\dfrac{4}{25}

B

\dfrac{2}{3}

C

\dfrac{2}{25}

D

\dfrac{4}{75}

Correct Answer

Option D

Solution

To solve this problem, we need to calculate the probability of two independent events occurring in succession: the first marble drawn is red, and the second marble drawn is white.

Since the drawing is with replacement, the number of marbles of each color remains the same for both draws.

The total number of marbles in the box is the sum of red, white, blue, and orange marbles:

\text{Total marbles} = 10 (\text{red}) + 30 (\text{white}) + 20 (\text{blue}) + 15 (\text{orange}) = 75.

The probability of drawing a red marble in the first draw is the number of red marbles divided by the total number of marbles:

P(\text{First is red}) = \frac{10}{75}.

Since the marble is replaced, the probability of drawing a white marble in the second draw remains as the number of white marbles divided by the total number of marbles:

P(\text{Second is white}) = \frac{30}{75}.

The probability of both independent events occurring in succession (drawing a red marble first and then a white marble) is the product of their individual probabilities:

P(\text{First is red and second is white}) = P(\text{First is red}) \times P(\text{Second is white}) = \frac{10}{75} \times \frac{30}{75}.

Now, let's calculate this probability:

P(\text{First is red and second is white}) = \frac{10 \times 30}{75 \times 75} = \frac{300}{5625}= \frac{4}{75}.

Therefore, the correct answer is Option D

\frac{4}{75}.

Q132

A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is

A

\dfrac{5}{11}

B

\dfrac{5}{6}

C

\dfrac{1}{6}

D

\dfrac{6}{11}

Correct Answer

Option A

Solution

Required probability $=$

\begin{aligned} & \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^3 \times \frac{1}{6}+\left(\frac{5}{6}\right)^5 \times \frac{1}{6}+\ldots . . \\ & =\frac{1}{6} \times \frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{11} \end{aligned}

Q133

An integer is chosen at random from the integers

1,2,3, \ldots, 50

. The probability that the chosen integer is a multiple of atleast one of 4, 6 and 7 is

A

\dfrac{8}{25}

B

\dfrac{9}{50}

C

\dfrac{14}{25}

D

\dfrac{21}{50}

Correct Answer

Option D

Solution

Given set

=\{1,2,3, \ldots \ldots . .50\}

\mathrm{P}(\mathrm{A})=

Probability that number is multiple of 4

\mathrm{P(B)}=

Probability that number is multiple of 6

\mathrm{P}(\mathrm{C})=

Probability that number is multiple of 7 Now,

\mathrm{P}(\mathrm{A})=\frac{12}{50}, \mathrm{P}(\mathrm{B})=\frac{8}{50}, \mathrm{P}(\mathrm{C})=\frac{7}{50}

again

\begin{aligned} & \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{4}{50}, \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\frac{1}{50}, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{1}{50} \\ & \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0 \end{aligned}

Thus

\begin{aligned} P(A & \cup B \cup C)=\frac{12}{50}+\frac{8}{50}+\frac{7}{50}-\frac{4}{50}-\frac{1}{50}-\frac{1}{50}+0 \\ & =\frac{21}{50} \end{aligned}

Q134

If an unbiased dice is rolled thrice, then the probability of getting a greater number in the

i^{\text{th }}

roll than the number obtained in the

(i-1)^{\text{th }}

roll,

i=2,3

, is equal to

A 5/54

B 2/54

C 1/54

D 3/54

Correct Answer

Option A

Solution

Let's denote the outcomes of the three rolls as

X_1

,

X_2

, and

X_3

, where

X_i

represents the number obtained in the

i^{\text{th}}

roll. We are looking for the probability that:

X_2 > X_1 \text{ and } X_3 > X_2

The total number of outcomes when rolling a dice three times is:

6^3 = 216

Let's count the favorable outcomes.

For each satisfying outcome, we must ensure that both inequalities are adhered to.

Consider the possible sequences where every subsequent roll's number is higher than the previous roll's number.

These sequences are: (1, 2, 3) (1, 2, 4) (1, 2, 5) (1, 2, 6) (1, 3, 4) (1, 3, 5) (1, 3, 6) (1, 4, 5) (1, 4, 6) (1, 5, 6) (2, 3, 4) (2, 3, 5) (2, 3, 6) (2, 4, 5) (2, 4, 6) (2, 5, 6) (3, 4, 5) (3, 4, 6) (3, 5, 6) (4, 5, 6) Clearly, there are 20 such favorable outcomes.

Hence, the probability is given by:

\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{20}{216} = \frac{5}{54}

Therefore, the correct option is: Option A:

\frac{5}{54}

Q135

Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn

\mathrm{A}

is :

A

\dfrac{4}{17}

B

\dfrac{5}{16}

C

\dfrac{5}{18}

D

\dfrac{7}{18}

Correct Answer

Option C

Solution

Let's denote the events as follows : Let

A_1, A_2,

and

A_3

be the events that urns A, B, and C are chosen, respectively. Let

B

be the event that a black ball is drawn.

We need to find the probability that the chosen urn is A given that a black ball is drawn, which is

P(A_1|B)

. Using Bayes' theorem, we have :

P(A_1|B) = \frac{P(B|A_1)P(A_1)}{P(B)}

First, we calculate each term individually : 1. The probability of choosing any urn, since they are chosen at random :

P(A_1) = P(A_2) = P(A_3) = \frac{1}{3}

2. The probability of drawing a black ball from each urn : Urn A:

P(B|A_1) = \frac{5}{12}

Urn B:

P(B|A_2) = \frac{7}{12}

Urn C:

P(B|A_3) = \frac{6}{12} = \frac{1}{2}

3. The total probability of drawing a black ball,

P(B)

:

P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + P(B|A_3)P(A_3)

P(B) = \frac{5}{12} \cdot \frac{1}{3} + \frac{7}{12} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3}

P(B) = \frac{5}{36} + \frac{7}{36} + \frac{6}{36}

P(B) = \frac{18}{36} = \frac{1}{2}

Now, substitute these into Bayes' theorem :

P(A_1|B)= \frac{P(B|A_1)P(A_1)}{P(B)}

P(A_1|B) = \frac{\frac{5}{12} \cdot \frac{1}{3}}{\frac{1}{2}}

P(A_1|B) = \frac{\frac{5}{36}}{\frac{1}{2}}

P(A_1|B) = \frac{5}{18}

Therefore, the probability that the black ball drawn is from urn A is option C :

\frac{5}{18}

Q136

There are three bags

X, Y

and

Z

. Bag

X

contains 5 one-rupee coins and 4 five-rupee coins; Bag

Y

contains 4 one-rupee coins and 5 five-rupee coins and Bag

Z

contains 3 one-rupee coins and 6 five-rupee coins. A bag is selected at random and a coin drawn from it at random is found to be a one-rupee coin. Then the probability, that it came from bag

\mathrm{Y}

, is :

A

\dfrac{1}{2}

B

\dfrac{1}{3}

C

\dfrac{5}{12}

D

\dfrac{1}{4}

Correct Answer

Option B

Solution

To solve this problem, we use Bayes' theorem. Let's define the events:

X

: Selecting bag

X

Y

: Selecting bag

Y

Z

: Selecting bag

Z

A

: Drawing a one-rupee coin We are given that a bag is selected at random, so the probabilities for choosing any of the bags are:

P(X) = P(Y) = P(Z) = \frac{1}{3}

Next, we need the probability of drawing a one-rupee coin from each bag: From bag

X

:

P(A|X) = \frac{5}{5+4} = \frac{5}{9}

From bag

Y

:

P(A|Y) = \frac{4}{4+5} = \frac{4}{9}

From bag

Z

:

P(A|Z) = \frac{3}{3+6} = \frac{3}{9} = \frac{1}{3}

We need to find the probability that the coin came from bag

Y

given that a one-rupee coin was drawn, i.e., we need

P(Y|A)

. Using Bayes' theorem:

P(Y|A) = \frac{P(A|Y) \cdot P(Y)}{P(A)}

To find

P(A)

, the total probability of drawing a one-rupee coin can be calculated as follows:

P(A) = P(A|X) \cdot P(X) + P(A|Y) \cdot P(Y) + P(A|Z) \cdot P(Z)

Substituting the values:

P(A) = \left(\frac{5}{9} \cdot \frac{1}{3}\right) + \left(\frac{4}{9} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right)

Calculating the above, we get:

P(A) = \frac{5}{27} + \frac{4}{27} + \frac{1}{9}

Note that

\frac{1}{9} = \frac{3}{27}

, so:

P(A) = \frac{5}{27} + \frac{4}{27} + \frac{3}{27} = \frac{12}{27} = \frac{4}{9}

Now, substituting back into Bayes' theorem:

P(Y|A) = \frac{\left(\frac{4}{9}\right) \cdot \left(\frac{1}{3}\right)}{\frac{4}{9}} = \frac{4}{9} \cdot \frac{1}{3} \cdot \frac{9}{4} = \frac{1}{3}

Hence, the probability that the coin came from bag

Y

is: Option B:

\frac{1}{3}

Q137

Let the sum of two positive integers be 24 . If the probability, that their product is not less than

\dfrac{3}{4}

times their greatest possible product, is

\dfrac{m}{n}

, where

\operatorname{gcd}(m, n)=1

, then

n

-

m

equals

A 10

B 11

C 9

D 8

Correct Answer

Option A

Solution

Take two numbers as

a

and

b

a+b=24

For product to be maximum

\begin{aligned} & \frac{a+b}{2} \geq \sqrt{a b} \\ & 144>a b \end{aligned}

Maximum product is 144 Now,

a b \geq \frac{3}{4} \cdot 144=108

Sample space

=\{(23,1),(22,2), \ldots\}

Integer points on line in shaded region

\begin{aligned} & \{(6,18),(7,17),(8,16), \ldots(18,6)\} \\ & P(E)=\frac{n(E)}{n(S)}=\frac{13}{23}=\frac{m}{n} \Rightarrow n-m=10 \end{aligned}

Q138

The coefficients

a, b, c

in the quadratic equation

a x^2+b x+c=0

are chosen from the set

\{1,2,3,4,5,6,7,8\}

. The probability of this equation having repeated roots is :

A

\dfrac{1}{128}

B

\dfrac{1}{64}

C

\dfrac{3}{256}

D

\dfrac{3}{128}

Correct Answer

Option B

Solution

Given quadratic equation is

a x^2+b x+c=0 \text{ where } a, b, c \in\{1,2,3, \ldots, 8\}

For repeated roots,

\begin{aligned} & b^2-4 a c=0 \\ & \Rightarrow b^2=4 a c \end{aligned}

\Rightarrow a c

must be a perfect square

(a, c) \in\{(1,1),(1,4),(2,2),(2,8),(3,3),(4,1),(4,4),(5,5),(6,6),(7,7),(8,2),(8,8)\}

Corresponding

b

must lie in set

\{1,2,3, \ldots 8\}

\begin{aligned} & (a, b, c) \in\{(1,2,1),(1,2,4),(2,4,2),(2,8,8) \\ & (3,6,3),(4,4,1),(4,8,4),(8,8,2)\} \\ & \therefore \text{ probability }=\frac{8}{8^3} \\ & =\frac{1}{64} \\ & \end{aligned}

Q139

The coefficients

\mathrm{a}, \mathrm{b}, \mathrm{c}

in the quadratic equation

\mathrm{a} x^2+\mathrm{bx}+\mathrm{c}=0

are from the set

\{1,2,3,4,5,6\}

. If the probability of this equation having one real root bigger than the other is p, then 216p equals :

A 38

B 76

C 57

D 19

Correct Answer

Option A

Solution

Equation is

a x^2+b x+c=0

\mathrm{D}>0

[for roots to be real & distinct]

\Rightarrow b^2-4 a c>0

For

b

\begin{aligned} & \text { For } b=3 \Rightarrow a c For

b=4 \Rightarrow a c

(a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3)\} \Rightarrow 5 \text { cases }

For

b=5 \Rightarrow a c

\begin{aligned} & (a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3),(2,2), \\ & (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6), \\ & (6,1)\}=14 \text{ cases } \end{aligned}

For

b=6 \Rightarrow a c

\begin{aligned} & (a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3),(2,2), \\ & (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6), \\ & (6,1),(2,4),(4,2)\}=16 \text { cases } \end{aligned}

Total cases

=3+5+14+16=38

cases

\Rightarrow

Probability,

p=\frac{38}{216}

\Rightarrow 216 p=38$$

Q140

If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is :

A

\dfrac{18}{25}

B

\dfrac{12}{25}

C

\dfrac{6}{25}

D

\dfrac{4}{25}

Correct Answer

Option B

Solution

We have 3 letters and 5 addresses, where 3 letters are posted to exactly 2 addresses.

First, we will select 2 addresses in

{ }^5 C_2

ways. Now, 3 letters can be posted to exactly 2 addresses in 6 ways.

\begin{aligned} \therefore \text{ Probability }=\frac{{ }^5 C_2 \times 6}{5^3} & \\ & =\frac{60}{125}=\frac{12}{25} \end{aligned}