Probability — Page 18 | JEE Mathematics

Q171

An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is :

A

{2 \over 7}

B

{1 \over 21}

C

{1 \over 23}

D

{1 \over 3}

Correct Answer

Option A

Solution

Out of nine balls three balls can be chosen =

{}^9{C_3}

ways $\therefore$ Sample space =

{}^9{C_3}

=

{{9!} \over {3!6!}}

=

{{9 \times 8 \times 7} \over 6}

= 84 According to the question, all three ball should be different.

So out of 3 red balls 1 is chosen and out of 4 blue 1 is chosen and out of 2 green 1 is chosen.

$\therefore$ Total cases =

{}^3{C_1} \times {}^4{C_1} \times {}^2{C_1}

= 3 $\times$ 4 $\times$ 2 = 24 $\therefore$ Probability =

{{24} \over {84}}

=

{{2} \over {7}}

Q172

An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is :

A

{{21} \over {49}}

B

{{27} \over {49}}

C

{{26} \over {49}}

D

{{32} \over {49}}

Correct Answer

Option D

Solution

5 Red and 2 green balls P(one red ball) =

{5 \over 7}

P(one green ball) =

{2 \over 7}

Case I : If drawn ball is green than a red ball is added

\left( \begin{array}{ll}{6{\mathop{\rm Re}\nolimits} d} \\ {1\,Green} \end{array} \right)

P (red ball) =

{6 \over 7}

Case II : If drawn ball is red than a green ball is added

\left( \begin{array}{ll}{4{\mathop{\rm Re}\nolimits} d} \\ {3\,Green} \end{array} \right)

P (red ball) =

{4 \over 7}

P (2nd red ball) =

{5 \over 7}

\times {4 \over 7} + {2 \over 7} \times {6 \over 7}

=

{{32} \over {49}}

Q173

A

and

B

are events such that

P\left( {A \cup B} \right) = 3/4

,

P\left( {A \cap B} \right) = 1/4,

P\left( {\overline A } \right) = 2/3

then

P\left( {\overline A \cap B} \right)

is :

A

5/12

B

3/8

C

5/8

D

1/4

Correct Answer

Option A

Solution

Given

P\left( {A \cup B} \right) = 3/4

,

P\left( {A \cap B} \right) = 1/4,

P\left( {\overline A } \right) = 2/3

We know,

P\left( A \right)

= 1 -

P\left( {\overline A } \right)

$\therefore$

P\left( A \right)

= 1 -

{2 \over 3}

=

{1 \over 3}

We know

P\left( {A \cup B} \right)

=

P\left( A \right)

+

P\left( B \right)

-

P\left( {A \cap B} \right)

$\Rightarrow$

{3 \over 4}

=

{1 \over 3}

+

P\left( B \right)

-

{1 \over 3}

$\Rightarrow$ 1 =

{1 \over 3}

+

P\left( B \right)

$\Rightarrow$

P\left( B \right)

=

{2 \over 3}

We know

P\left( {\overline A \cap B} \right)

=

P\left( B \right)

-

P\left( {A \cap B} \right)

So

P\left( {\overline A \cap B} \right)

=

{2 \over 3}

-

{1 \over4}

=

{5 \over 12}

Q174

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is :

A

{2 \over 9}

B

{1 \over 9}

C

{8 \over 9}

D

{7 \over 9}

Correct Answer

Option B

Solution

Person 1st has three options to apply.

Similarly, person 2nd has three options t apply and person 3rd has three options to apply.

Total cases = 33 Now, favourable cases = 3 (An either all has applied for house 1 or 2 or 3) So, probability

= {3 \over {{3^3}}} = {1 \over 9}

.

Q175

Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is :

A

{{2 \over 5}}

B

{{4 \over 5}}

C

{{3 \over 5}}

D

{{1 \over 5}}

Correct Answer

Option A

Solution

X : Mr. A selected wining horse

\overline X

: Mr.

A did not select wining horse Mr.

A selected two horses, now probability of not wining the first horse which Mr.

A choses =

{4 \over 5}

And probability of not wining the second horse also which Mr. A choses =

{3 \over 4}

(Here Mr. A out of remaining 4 horses choses one horse among 3 horses which did not win) $\therefore$

P(\overline X ) = {4 \over 5} \times {3 \over 4}

$\therefore$

P(X) = 1 - P(\overline X )

= 1 - {4 \over 5} \times {3 \over 4}

= 1 - {3 \over 5}

= {2 \over 5}

Q176

A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let

X

denote the number of defective pens. Then the variance of

X

is

A

\dfrac{11}{15}

B

\dfrac{2}{15}

C

\dfrac{3}{5}

D

\dfrac{28}{75}

Correct Answer

Option D

Solution

.tg .tg $X$ $P(X)$ $XP(X)$ $\left(X_i-\mu\right)^2$ $P_i X\left(X_i-\mu\right)^2$ $X=0$ $\dfrac{{ }^7 C_2}{{ }^{10} C_2}$ 0 $\left(0-\dfrac{3}{5}\right)^2$ $\dfrac{7}{15}\left(\dfrac{9}{25}\right)$ $X=1$ $\dfrac{{ }^7 C_1{ }^3 C_1}{{ }^{10} C_2}$ $\dfrac{7}{15}$ $\left(1-\dfrac{3}{5}\right)^2$ $\dfrac{7}{15}\left(\dfrac{4}{25}\right)$ $x=2$ $\dfrac{{ }^7 C_0{ }^3 C_2}{{ }^{10} C_2}$ $\dfrac{2}{15}$ $\left(2-\dfrac{3}{5}\right)^2$ $\dfrac{2}{15}\left(\dfrac{49}{25}\right)$

\begin{aligned} & \mu=\sum X_i P\left(X_i\right)=0+\frac{7}{15}+\frac{2}{15}=\frac{3}{5} \\ & \text{ Variance }(X)= \\ & \sum P_i\left(X_i-\mu\right)^2=\frac{7}{15}\left(\frac{9}{25}\right)+\frac{7}{15}\left(\frac{4}{25}\right)+\frac{2}{15}\left(\frac{49}{25}\right)=\frac{28}{75} \end{aligned}

Q177

If the mean of the following probability distribution of a radam variable

\mathrm{X}

: .tg .tg

\mathrm{X}

0 2 4 6 8

\mathrm{P(X)}

a

2a

a+b

2b

3b

is

\dfrac{46}{9}

, then the variance of the distribution is

A

\dfrac{581}{81}

B

\dfrac{566}{81}

C

\dfrac{151}{27}

D

\dfrac{173}{27}

Correct Answer

Option B

Solution

.tg .tg

X

0 2 4 6 8

P(X)

a

2a

a+b

2b

3b

\begin{aligned} & \text{ Mean }=\sum x_i P\left(x_i\right) \\ & \frac{46}{9}=4 a+4 a+4 b+12 b+24 b \\ & \frac{46}{9}=8 a+40 b \\ & \frac{23}{9}=4 a+20 b \\ & 36 a+180 b=23 \quad \text{.... (1)} \end{aligned}

Sum of probability is 1

\Rightarrow 4 a+6 b=1 \quad \text{... (2)}

\begin{aligned} & \text{ Solving (1) and (2) } \\ & a=\frac{1}{12}, b=\frac{1}{9} \\ & \sigma^2=\sum x_i^2 P\left(x_i\right)-\left(\sum x_i P\left(x_i\right)\right)^2 \\ & =4 \times 2 a+16(a+b)+36(2 b)+64(3 b)-\left(\frac{46}{9}\right)^2 \\ & =8(a+2(a+b)+9 b+24 b)-\left(\frac{46}{9}\right)^2 \\ & =8(3 a+35 b)-\left(\frac{46}{9}\right)^2 \\ & =8\left(\frac{3}{12}+\frac{35}{9}\right)-\left(\frac{46}{9}\right)^2 \\ & =8\left(\frac{149}{36}\right)-\left(\frac{46}{9}\right)^2=\frac{566}{81} \\ \end{aligned}

Q178

Four numbers are chosen at random (without replacement) from the set

\left\{ {1,2,3,....20} \right\}.

Statement - 1: The probability that the chosen numbers when arranged in some order will form an AP is

{1 \over {85}}.

Statement - 2: If the four chosen numbers form an AP, then the set of all possible values of common difference is

\left( { \pm 1, \pm 2, \pm 3, \pm 4, \pm 5} \right).

A Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1.

B Statement - 1 is true, Statement - 2 is false.

C Statement - 1 is false, Statement -2 is true.

D Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1.

Correct Answer

Option B

Solution

Four numbers can be chosen

{}^{20}{C_4}

ways.

When common difference d = 1 then the possible sets are (1, 2, 3, 4) (2, 3, 4, 5) (3, 4, 5, 6) ................. (17, 18, 19, 20) = 17 sets So when d = 1 then 17 different AP's are possible with 4 numbers.

Now let's create a table of all possible sets - .tg .tg Common Difference (d) Possible Sets No of AP d = 1 (1, 2, 3, 4) (2, 3, 4, 5) (3, 4, 5, 6) ................. (17, 18, 19, 20) 17 d = 2 (1, 3, 5, 7) (2, 4, 6, 8) (3, 5, 7, 9) ................. (14, 16, 18, 20) 14 d = 3 (1, 4, 7, 10) (2, 5, 8, 11) (3, 6, 9, 12) ................. (11, 14, 17, 20) 11 d = 4 (1, 5, 9, 13) (2, 6, 10, 14) (3, 7, 11, 15) ................. (8, 12, 16, 20) 8 d = 5 (1, 6, 11, 16) (2, 7, 12, 17) (3, 8, 13, 18) (4, 9, 14, 19) (5, 10, 15, 20) 5 d = 6 (1, 7, 13, 19) (2, 8, 14, 20) 2 $\therefore$ Total no of AP = 17 + 14 + 11 + 8 + 5 + 2 = 57 $\therefore$ Required probability =

{{57} \over {{}^{20}{C_4}}}

=

{1 \over {85}}

$\therefore$ Statement - 1: is true. $\therefore$ Statement - 2: is false as common difference can also be

\pm 6

.

Q179

Let S = {1, 2, . . . . . ., 20}. A subset B of S is said to be "nice", if the sum of the elements of B is 203. Then the probability that a randonly chosen subset of S is "nice" is :

A

{5 \over {{2^{20}}}}

B

{7 \over {{2^{20}}}}

C

{4 \over {{2^{20}}}}

D

{6 \over {{2^{20}}}}

Correct Answer

Option A

Solution

We can solve this problem by counting the number of "nice" subsets in the set S = {1, 2, $\ldots$ , 20 }, and then dividing that number by the total number of possible subsets of S.

The sum of all elements in S is : 1 + 2 + $\ldots$ + 20 = $\dfrac{{20 \times 21}}{2}$ = 210 Since a "nice" subset must sum to 203, the elements not in the subset must sum to 210 - 203 = 7.

Now we need to find the ways to make the sum of 7 using the elements of S.

The combinations are : 1.

7 2.

1 + 6 3.

2 + 5 4.

3 + 4 5.

1 + 2 + 4 6.

1 + 3 + 3(This doesn't work since 3 is repeated) 7.

2 + 2 + 3(This doesn't work since 2 is repeated) So, there are 5 "nice" subsets.

Since the set S has 20 elements, there are $2^{20}$ possible subsets (including the empty set and the set itself).

The probability of randomly choosing a "nice" subset is therefore :

Q180

If A and B are any two events such that P(A) =

{2 \over 5}

and P (A

\cap

B) =

{3 \over {20}}

, hen the conditional probability, P(A

\left| {} \right.

(A'

\cup

B')), where A' denotes the complement of A, is equal to :

A

{1 \over 4}

B

{5 \over 17}

C

{8 \over 17}

D

{11 \over 20}

Correct Answer

Option B

Solution

P\left( {{A \over {A' \cup B'}}} \right)

= {{P\left[ {A \cap \left( {A' \cup B'} \right)} \right]} \over {P\left( {A' \cap B'} \right)}}

= {{P\left[ {\left( {A \cap A'} \right) \cup \left( {A \cap B'} \right)} \right]} \over {P\left( {A \cap B} \right)'}}

\left[ \, \right.

As

\left. {\left( {A \cap B} \right)' = A' \cap B'} \right]

= {{P\left( {A \cap B'} \right)} \over {P\left( {A \cap B} \right)'}}

As

A \cap A' = \phi

$\therefore$

\phi \cup \left( {A \cap B'} \right) = A \cap B'

= {{P\left( A \right) - P\left( {A \cap B} \right)} \over {1 - P\left( {A \cap B} \right)}}

\left[ \, \right.

As

A \cap B' = A - \left( {A \cap B} \right)

and

\left. {\left( {A \cap B} \right)' = 1 - \left( {A \cap B} \right)} \right]

Given

P\left( A \right) = {2 \over 5}

and

P\left( {A \cap B} \right) = {3 \over {20}}

= {{{2 \over 5} - {3 \over {20}}} \over {1 - {3 \over {20}}}}

= {{{{8 - 3} \over {20}}} \over {{{17} \over {20}}}}

= {5 \over {17}}