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Q:

In a charity show tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random (blindly) will have a number with a hundredth digit of 2?

A) 0.25	B) 0.5
C) 0.75	D) 0.40

Answer & Explanation Answer: D) 0.40

Explanation:

250 numbers between 101 and 350 i.e. n(S)=250

n(E)=100th digits of 2 = 299−199 = 100

P(E)= n(E)/n(S) = 100/250 = 0.40

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Q:

Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King

A) 1/52	B) 1/26
C) 1/13	D) 1/2

Answer & Explanation Answer: B) 1/26

Explanation:

Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
=P(A)+P(B)

= $[\frac{12 C_{1} * 1 C_{1}}{52 C_{2}}] + [\frac{13 C_{1} * 3 C_{1}}{52 C_{2}}]$

= $[12 * \frac{2}{52 * 51}] + [13 * 3 * \frac{2}{52 * 51}]$ = $\frac{24 + 78}{52 * 51}$ = $\frac{1}{26}$

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Q:

The probability of success of three students X, Y and Z in the one examination are 1/5, 1/4 and 1/3 respectively. Find the probability of success of at least two.

A) 1/4	B) 1/2
C) 1/6	D) 1/3

Answer & Explanation Answer: C) 1/6

Explanation:

P(X) = $\frac{1}{5}$ , P(Y) = $\frac{1}{4}$ , P(Z) = $\frac{1}{3}$

Required probability:

= [ P(A)P(B){1−P(C)} ] + [ {1−P(A)}P(B)P(C) ] + [ P(A)P(C){1−P(B)} ] + P(A)P(B)P(C)

= $(\frac{1}{4} * \frac{1}{3} * \frac{4}{5}) + (\frac{3}{4} * \frac{1}{3} * \frac{1}{5}) + (\frac{2}{3} * \frac{1}{4} * \frac{1}{5}) + (\frac{1}{4} * \frac{1}{3} * \frac{1}{5})$

= $\frac{4}{60} + \frac{3}{60} + \frac{2}{60} + \frac{1}{60}$ = $\frac{10}{60}$ = $\frac{1}{6}$

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Q:

You toss a coin AND roll a die. What is the probability of getting a tail and a 4 on the die?

A) 1/2	B) 1/12
C) 2/3	D) 3/4

Answer & Explanation Answer: B) 1/12

Explanation:

Probability of getting a tail when a single coin is tossed =12
Probability of getting 4 when a die is thrown =16

Required probability =(12)×(16)
= 1/12

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Q:

A bag contains 7 green and 5 black balls. Three balls are drawn one after the other. The probability of all three balls being green, if the balls drawn are not replaced will be:

A) 123/897	B) 23/67
C) 7/44	D) 12/45

Answer & Explanation Answer: C) 7/44

Explanation:

Here, n(E) = $7 C_{1} \times 5 C_{1} \times 5 C_{1}$

And, n(S) = $12 C_{1} * 11 C_{1} * 10 C_{1}$

P(S) = $\frac{7 * 6 * 5}{12 * 11 * 10}$ = 7/44

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Q:

Out of 17 applicants 8 boys and 9 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a girl.

A) 19/34	B) 5/4
C) 20/34	D) 25/34

Answer & Explanation Answer: D) 25/34

Explanation:

The events of selection of two person is redefined as first is a girl and second is a boy or first is boy and second is a girl or first is a girl and second is a girl.

So the required probability:
= $(\frac{8}{17} * \frac{9}{16}) + (\frac{9}{17} * \frac{8}{16}) + (\frac{8}{17} * \frac{7}{16})$

= $\frac{9}{34} + \frac{9}{34} + \frac{7}{34}$
= $\frac{25}{34}$

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Q:

Series of interface points that allow other computers to communicate with the other layers of network protocol stack is called SAP

A) TRUE

B) FALSE

Answer & Explanation Answer: A) TRUE

Explanation:

Workspace

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Subject: Network Administration
Job Role: Network Engineer

Q:

From a pack of 52 cards, 3 cards are drawn. What is the probability that one is ace, one is queen and one is jack?

A) 19/5525	B) 16/5525
C) 17/5525	D) 7/5525

Answer & Explanation Answer: B) 16/5525

Explanation:

Required probability:

$\frac{4 C_{1} * 4 C_{1} * 4 C_{1}}{52 C_{3}} = \frac{4 * 4 * 4}{22100} = \frac{16}{5525}$

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Filed Under: Probability

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