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Q:

What is the state of the processor, when a process is waiting for some event to occur?

Answer

Waiting state

Workspace

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Subject: Operating Systems

Q:

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) 2/9	B) 1/9
C) 8/9	D) 7/9

Answer & Explanation Answer: B) 1/9

Explanation:

One person can select one house out of 3= $3 C_{1}$ ways =3.

Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9.

Therefore, probability that all thre apply for the same house is 1/9

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Filed Under: Probability

Q:

Find the cost of 96 shares of Rs. 10 each at $\frac{3}{4}$ discount, brokerage being 1/4 per share.

A) 912	B) 921
C) 920	D) 900

Answer & Explanation Answer: A) 912

Explanation:

Cost of 1 share = $R s . [(10 - \frac{3}{4}) + \frac{1}{4}] = R s \frac{19}{2}$

Cost of 96 shares = $R s . (\frac{19}{2} * 96)$ = Rs. 912.

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Filed Under: Stocks and Shares

Q:

Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?

A) 52/221	B) 55/190
C) 55/221	D) 19/221

Answer & Explanation Answer: C) 55/221

Explanation:

We have n(s) = $52 C_{2}$ 52 = 52*51/2*1= 1326.

Let A = event of getting both black cards

B = event of getting both queens

A∩B = event of getting queen of black cards

n(A) = $\frac{52 * 51}{2 * 1}$ = $26 C_{2}$ = 325, n(B)= $\frac{26 * 25}{2 * 1}$ = 4*3/2*1= 6 and n(A∩B) = $4 C_{2}$ = 1

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and

P(A∩B) = n(A∩B)/n(S) = 1/1326

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

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Filed Under: Probability

Q:

Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6.

A) 7/18	B) 14/35
C) 8/18	D) 7/35

Answer & Explanation Answer: A) 7/18

Explanation:

Clearly, n(S) = 6 x 6 = 36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.
Then,E = {(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)}
n(E) = 14.
Hence, P(E) = n(E)/n(S) = 14/36 = 7/18

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Filed Under: Probability

Q:

In a simultaneous throw of pair of dice. Find the probability of getting the total more than 7.

A) 1/2	B) 5/12
C) 7/15	D) 3/12

Answer & Explanation Answer: B) 5/12

Explanation:

Here n(S) = (6 x 6) = 36

Let E = event of getting a total more than 7
= {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.

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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:

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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