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Q:

I. A. B, C. D, E and F are six members of a family.

II. One couple has parents and their children in the family.

III. A is the son of C and E is the daughter of A.

IV. D is the daughter of F who is the mother of E.

Which of the following pairs is the parents of the couple ?

A) AB	B) BC
C) AF	D) CF

Answer & Explanation Answer: B) BC

Explanation:

E is the daughter of A and F is the mother of E So, A is the father of E and hence the husband of F. Now,

D is the daughter of F So. D and E are the daughters of A and F Also. A is the son of C.

Now, only B remains Thus. B and C are the parents of A.

Clearly. B and C are the parents of the couple

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Filed Under: Blood Relations

Q:

What is relative path and absolute path?

Answer

Absolute path-- Exact path from root directory.

Relative path-- Relative to the current path.

Workspace

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Subject: Operating Systems

Q:

If a box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains no defective bulbs.

A) 5/12	B) 7/12
C) 3/14	D) 1/12

Answer & Explanation Answer: D) 1/12

Explanation:

Total number of elementary events = $10 C_{5}$

Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is $7 C_{5}$ .

So,required probability = $7 C_{5}$ / $10 C_{5}$ = 1/12.

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Filed Under: Probability

Q:

A box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains exactly one defective bulb.

A) 5/12	B) 7/12
C) 3/14	D) 1/12

Answer & Explanation Answer: A) 5/12

Explanation:

Total number of elementary events = $10 C_{5}$

Number of ways of selecting exactly one defective bulb out of 3 and 4 non-defective out of 7 is $3 C_{1} * 7 C_{4}$

So,required probability = $3 C_{1} * 7 C_{4}$ / $10 C_{5}$ = 5/12.

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Filed Under: Probability

Q:

A bag contains 50 tickets numbered 1,2,3,4......50 of which five are drawn at random and arranged in ascending order of magnitude.Find the probability that third drawn ticket is equal to 30.

A) 551/15134	B) 1/2
C) 552/15379	D) 1/9

Answer & Explanation Answer: A) 551/15134

Explanation:

Total number of elementary events = $50 C_{5}$
Given,third ticket =30

=> first and second should come from tickets numbered 1 to 29 = $29 C_{2}$ ways and remaining two in $20 C_{2}$ ways.

Therfore,favourable number of events = $29 C_{2} * 20 C_{2}$

Hence,required probability = $29 C_{2} * 20 C_{2}$ $/ 50 C_{5}$ =551 / 15134

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Filed Under: Probability

Q:

What is NETBIOS and NETBEUI

Answer

NETBIOS is a programming interface that allows I/O requests to be sent to and received from a remote computer and it hides the networking hardware from applications.

NETBEUI is NetBIOS extended user interface. A transport protocol designed by microsoft and IBM for the use on small subnets.

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Subject: Networking
Job Role: Network Engineer

Q:

Find the probability of selecting 2 woman when four persons are choosen at random from a group of 3 men, 2 woman and 4 children.

A) 1/5	B) 1/7
C) 1/6	D) 1/9

Answer & Explanation Answer: C) 1/6

Explanation:

Out of 9 persons,4 can be choosen in $9 C_{4}$ ways =126.

Favourable events for given condition = $2 C_{2} * 7 C_{2}$ = 21.

So,required probability = 21/126 =1/6.

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Filed Under: Probability

Q:

A word consists of 9 letters; 5 consonants and 4 vowels.Three letters are choosen at random. What is the probability that more than one vowel will be selected ?

A) 13/42	B) 17/42
C) 5/42	D) 3/14

Answer & Explanation Answer: B) 17/42

Explanation:

3 letters can be choosen out of 9 letters in $9 C_{3}$ ways.

More than one vowels ( 2 vowels + 1 consonant or 3 vowels ) can be choosen in $(4 C_{2} * 5 C_{1}) + 4 C_{3}$ ways

Hence,required probability = $\frac{(4 C_{2} * 5 C_{1}) + 4 C_{3}}{9 C_{3}}$ = 17/42

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Filed Under: Probability

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