Option 3 : 39 bits/sec

__Concept:__

Information associated with the event is “inversely” proportional to the probability of occurrence.

Entropy: The average amount of information is called the “Entropy”.

\(H = \;\mathop \sum \limits_i {P_i}{\log _2}\left( {\frac{1}{{{P_i}}}} \right)\;bits/symbol\)

Rate of information = r.H

__Calculation:__

Given: r = 24 outcomes/sec

\(P_1=\frac{1}{2}\), \(P_2=\frac{1}{ 4}\), \(P_3=\frac{1}{ 8}\) and \(P_4=\frac{1}{ 16}\)

\( H = \frac{1}{2}{\log _22} + \frac{1}{4}{\log_24};+ \frac{1}{8}{\log_28};+ \frac{1}{16}{\log_216};\)

H = 1.625 bits/outcome

∴ Rate of information = r.H

Rs = 24 x 1.625

Rs = 39 bits/sec

Option 2 : 7/8

A die is having six sides with six different numbering

Even number in the die are {2,4,6}

The odd number in the die are {1,3,5}

\(Probability~of~an~Event =\frac{Number~of~desired~Events }{Total~number~of~Events}\)

The probability of occurring an odd number is = 3/6 = 1/2

The probability of occurring an even number is = 3/6 = 1/2

When die is tossed three times and getting an odd number **at least once** is given by

p(at least one odd number) = 1 - p(no odd number in all three tosses)

\(p(at~least~one~odd~number) =1- \frac{1}{8}\)

p(no odd number in all three tosses) = p(even) × p(even) × p(even)

\(p(no~odd~number~in~all~three~tosses) = \frac{1}{2}×\frac{1}{2}×\frac{1}{2}\)

\(p(no~odd~number~in~all~three~tosses) = \frac{1}{8}\)

\(p(at~least~one~odd~number) =\frac{7}{8}\)

Hence **option (2) is correct **

Option 3 : 1/4

__Data:__

No. of Students in class = 30 students

30 students have to sit on 40 seats

__Calculation:__

Total seats = (5 row) × (8 seats/row) = 40 seats

Total no. of ways = ^{40}C_{30}

For 6^{th} seat of 5^{th} row to be empty, all 30 students have to sit on remaining 39 seats.

No. of favourable ways = ^{39}C_{30}

Probability (sixth seat in the fifth row will be empty) =

\(= \frac{{{\rm{NO}}.{\rm{\;of\;favourable\;ways}}}}{{{\rm{Total\;no}}.{\rm{\;of\;ways}}}}\)

= ^{39}C_{30} ÷ ^{40}C_{30}

= ¼

Option 4 : 41/70

**Concept:**

1^{st} purse contains 4 copper and 3 silver coins, so the probability of selecting 1 copper coin from it,

Thus, P (copper coin from 1^{st} purse) = \(\frac{^4C_1}{^7C_1}\) = \(\frac{4}{7}\)

2^{nd} purse contains 6 copper and 4 silver coins, so the probability of selecting 1 copper coin from it,

Thus, P (copper coin from 2^{nd} purse) = \(\frac{^6C_1}{^10C_1}\)= \(\frac{6}{{10}}\)

And, Probability of selecting any purse = \(\frac{1}{2}\)

Hence, the Probability of selecting a copper coin = \(\frac{1}{2} \times \left( {\frac{4}{7} + \frac{6}{{10}}} \right)\)

**Probability of selecting a copper coin = \(\frac{{41}}{{70}}\)**

Option 2 : 4/7

**Concept:**

P(E/F) = \(\frac{{{\rm{P}}\left( {{\rm{E}} ∩ {\rm{F}}} \right)}}{{{\rm{P}}\left( {\rm{F}} \right)}}\)

P(E⋂F) = P(E/F) × P(F)

P(E⋂F) = 0.3 P(F) ---(1)

Also, P(EUF) = P(E) + P(F) - P(E⋂F)

0.8 = 0.4 + P(F) - 0.3 P(F) (From equation (i))

0.4 = 0.7 P(F)

A box contains the following three coins.

I. A fair coin with head on one face and tail on the other face.

II. A coin with heads on both the faces.

III. A coin with tails on both the faces.

A coin is picked randomly from the box and tossed. Out of the two remaining coins in the box, one coin is then picked randomly and tossed. If the first toss results in a head, the probability of getting a head in the second toss is

Option 2 : \(\frac{1}{3}\)

__Concept__**:**

Conditional probability is defined as:

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

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

__Application__**:**

Let event A is defined as:

A = Getting head in the first toss

Event B is defined as:

B = Getting head in the second toss

According to the question, we need to find the probability of getting a head in the second toss when

Already a head has occurred in the first toss, i.e.So, the probability of getting head in the first toss will be:

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

P(A) = P(fair coin is selected) × P(Getting head) + P(double-headed coin is selected) × P(getting head)

\( = \frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times 1 = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}\)

\(P\left( A \right) = \frac{1}{2}\)

Now, finding the probability of getting head in both tosses will be:

\(P\left( {A \cap B} \right) = \mathop {\mathop {\underbrace {\begin{array}{*{20}{c}} {First\;toss}\\ {\left( {\frac{1}{3} \cdot \frac{1}{2}} \right)} \end{array}\;\;\begin{array}{*{20}{c}} {second\;toss}\\ {\left( {\frac{1}{2} \cdot 1} \right)} \end{array}}_{\begin{array}{*{20}{c}} {when\;first\;fair}\\ {coin\;is\;tossed} \end{array}}\;\;}\limits_{} }\limits_\; + \underbrace {\begin{array}{*{20}{c}} {First\;toss}\\ {\left( {\frac{1}{3} \cdot 1} \right)} \end{array}\;\;\begin{array}{*{20}{c}} {second\;toss}\\ {\left( {\frac{1}{2} \cdot \frac{1}{2}} \right)} \end{array}}_{\begin{array}{*{20}{c}} {When\;first\;double}\\ {headed\;coin\;is\;tossed} \end{array}}\)

\(P\left( {A \cap B} \right) = \frac{1}{{12}} + \frac{1}{{12}} = \frac{1}{6}\)

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

\(P\left( {\frac{B}{A}} \right) = \frac{{\frac{1}{6}}}{{\frac{1}{2}}} = \frac{1}{3}\)

__Concept:__

Independent events: If the occurrence of one event do not depend on the occurrence of another event, then both events are called independent events.

In case of independent events,

P(A/B) = P(B)

∴ P(A ∩ B) = P(A) ⋅ P(B)

Addition theorem of probability:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

**where A ∪ B = At least one of A or B**

Multiplication theorem of probability:

P(A ∩ B) = P(A/B) ⋅ P(B)

where A ∩ B = Simultaneous occurrence of A and B

__Calculation:__

__Given:__

P(F) = 0.3, P(G) = 0.4, P(F ⋂ G) = 0.2

Addition theorem of probability:

P(F ∪ G) = P(F) + P(G) – P(F ∩ G)

The probability of occurrence of at least one event,

P(F ∪ G) = 0.3 + 0.4 - 0.2 = 0.5

Option 1 : 1/625

The prime factors of 10 are 2 and 5.

The divisors of 10^{n} are of the form 2^{p}, 5^{q} when p and q take values from 0 to n.

Number of divisors of 10^{n} = (n + 1)^{2}

Number of divisors of 10^{99} = 10,000

Number of divisors of 10^{99} which are multiples of 10^{96} = Number of divisors of 10^{3} = 4 x 4

Required probability = 16/10,000 = 1/625

**Hence the correct answer is** 1/625.

Option 4 : 40, 6

**Concept:**

Mean = np

Variance = np (1-p)

where, p is the probability of defect

**Analysis:**

Mean of items = 0.1 × 400 = 40

And variance \(\sigma^2 = 40 \times 0.9 = 36\)

So standard deviation = √ variance = σ = √ 36 = 6

Hence, (d) is the correct choice.

Option 1 : 4/3

**Concept used**

We know that the probability of joint density \(\iint \)f_{xy}(xy)dxdy

**Calculation**

\(\mathop \iint \limits_{x = 0y = 0}^{x = 1y = 1} \) (x^{2} + Cy)dxdy = 1

After integration we get

⇒ (x^{3}/3)^{1}_{0} + C(y^{2}/2)^{1}_{0} = 1

⇒ (1/3) + (1/2)C = 1

⇒ (1/2)C = 1 – 1/3

⇒ C(1/2) = 2/3

⇒ C = 4/3

**∴ The value of C is 4/3**

Option 1 : pq + (1 - p) (1 - q)

Probability of faculty assembly of any computer=p

The probability that the testing process gives the correct result= q

Required probability= Probability of faulty assemble when it is tested correct or Probability of right assemble when it is tested incorrect.

=(pq)+(1-p)(1-q)

**Hence the correct answer is ***pq + (1 - p) (1 - q)*

Option 1 : 1/5

The correct answer is **option 1.**

**EXPLANATION:**

There are 5 cards (each carrying a distinct number from 1 to 5) in the deck.

- | 1,2 | 1,3 | 1,4 | 1,5 |

2,1 |
- | 2,3 | 2,4 | 2,5 |

3,1 | 3,2 |
- | 3,4 | 3,5 |

4,1 | 4,2 | 4,3 |
- | 4,5 |

5,1 | 5,2 | 5,3 | 5,4 |
- |

So arranging two cards among 5 cards, sample space will 5p3= 20 ways.

S → Total sample space.

n(S) = 20

E → Two cards are selected with the number on the first card being one higher than the number on the second card.

n(E)= {(2,1), (3,2), (4,3), (5,4) } = 4 ways

P(E)=\(\frac{n(E)}{n(S)}={4 \over 20 }= {1 \over 5}\)

**Hence the correct answer is** \(1 \over 5\).

Option 1 : 0.101

**Concept:**

When all the four rules are used simultaneously, the probability of overall Type-I error is given by :

\(P\left ( \frac{Overall}{Error} \right )\) = P(Error in at least of them)

**Calculation:**

**Given:**

The probability of Type-I error for the four rules are 0.005, 0.02, 0.03, and 0.05.

The probability of overall Type-I error is given by :

\(P\left ( \frac{Overall}{Error} \right )\) = P(Error in at least of them)

= 1 - P(Error in none of them)

= 1 - (0.995 \(\times\) 0.98 \(\times\)0.97 \(\times\) 0.95)

= 1 - 0.898

= 0.101

**∴ The probability of overall Type-I error when all the four rules are used simultaneously is 0.101**

Option 4 : \(\frac{1}{5}\)

**Given**

P(x) = {x/15 ; x = 1, 2, 3, ----5

{0 ; otherwise

**Calculation**

Probability P{1/2 < X < 5/2} that means, X = 1, 2 are only applicable

∴P(x = 1) = 1/15

∴ P(x = 2) = 2/15

⇒ P(x ) = p(x = 1) + p(x = 2)

⇒ p(x) = 1/15 + 2/15

**∴**** The probability of p(x) is 3/15 or 1/5**

The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is _____

Numbers which are divisible by any one of 2, 3, and 5

∴ Number of numbers from 1 to 100 which are divisible by 2 = 100/2 = 50

Number of numbers from 1 to 100 which are divisible by 3 but not by 2 = Numbers divisible by 3 - numbers divisible by 6

⇒ 99/3 - 96/6 = 17

Number of numbers from 1 to 100 which are divisible by 5 but not by 2 or 3 = Numbers divisible by 5 - numbers divisible by 10 - numbers divisible by 15 + numbers divisible by 30

⇒ 100/5 - 100/10 - 90/15 + 90/30

⇒ 20 - 10 - 6 + 3

⇒ 7

NOT divisible by 2, 3 or 5 = 100 – (50 + 17 + 7) = 26

\({\rm{Probability}} = \frac{{26}}{{100}} = {\rm{\;}}0.26\)

The security system at an IT office is composed of 10 computers of which exactly four are working. To check whether the system is functional, the officials inspect four of the computers picked at random (without replacement). The system is deemed functional if at least three of the four computers inspected are working. Let the probability that the system is deemed functional be denoted by p. Then 100p = _____________.

__Data:__

Total computer = 10

Computer working = 4

Computer not working = 6

P(W) = probability of working computer

P(N) = probability of not working computer

__Calculation:__

\({\rm{P}}\left( {\rm{W}} \right) = \frac{4}{{10}}{\rm{\;and\;P}}\left( {\rm{N}} \right) = \frac{6}{{10}}\)

At least 3 computers need to be working out of 4

Any 3 is working or 4 are working

The probability that the system is deemed functional without replacement

p = P(WWWN) + P(WWNW) + P(WNWW) + P(NWWW) + P(WWWW)

p = 4 × P(WWWN) + P(WWWW)

\({\rm{p}} = 4 \times \frac{4}{{10}} \times \frac{3}{9} \times \frac{2}{8} \times \frac{6}{{7{\rm{\;}}}} + \frac{4}{{10}} \times \frac{3}{9} \times \frac{2}{8} \times \frac{1}{7}\)

\({\rm{p}} = \frac{{600}}{{5040}}\)

\(100{\rm{p}} = 100 \times \frac{{600}}{{5040}} = 11.9047\)