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Q:

In a simultaneous throw of two dice, what is the probability of getting a total of 7 ?

A) 1/6	B) 1/4
C) 1/20	D) 3/4

Answer & Explanation Answer: A) 1/6

Explanation:

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Filed Under: Probability

Q:

In a simultaneous throw of two dice, what is the probability of getting a doublet ?

A) 1/6	B) 1/3
C) 4/7	D) 4/5

Answer & Explanation Answer: A) 1/6

Explanation:

In a simultaneous throw of two dice, n(S) = 6 x 6 = 36

Let E = event of getting a doublet = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

$P (E) = \frac{n (E)}{n (S)} = \frac{6}{36} = \frac{1}{6}$

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Q:

The value of $(\frac{1}{\log_{3} 60} + \frac{1}{\log_{4} 60} + \frac{1}{\log_{5} 60}) i s$

A) 0	B) 1
C) 5	D) 60

Answer & Explanation Answer: B) 1

Explanation:

=> $\log_{60} (3 * 4 * 5)$

=> $\log_{60} (60)$

= 1

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Filed Under: Logarithms
Exam Prep: Bank Exams
Job Role: Bank PO

Q:

$i f \log \frac{a}{b} + \log \frac{b}{a} = \log (a + b), t h e n$

A) a + b = 1	B) a - b = 1
C) a = b	D) ab=1

Answer & Explanation Answer: A) a + b = 1

Explanation:

$i f \log \frac{a}{b} + \log \frac{b}{a} = \log (a + b), t h e n$

$\log (a + b) = \log (\frac{a}{b} \times \frac{b}{a}) = \log 1$

so, a+b=1

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Filed Under: Logarithms
Exam Prep: Bank Exams
Job Role: Bank PO

Q:

What is the probability of getting at least one six in a single throw of three unbiased dice?

A) 1/36	B) 91/256
C) 13/256	D) 43/256

Answer & Explanation Answer: B) 91/256

Explanation:

Find the number of cases in which none of the digits show a '6'.

i.e. all three dice show a number other than '6', 5×5×5=125 cases.

Total possible outcomes when three dice are thrown = 216.

The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.

=216−125=91

The required probability = 91/256

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Q:

Four dice are thrown simultaneously. Find the probability that all of them show the same face.

A) 1/216	B) 1/36
C) 2/216	D) 4/216

Answer & Explanation Answer: A) 1/216

Explanation:

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:

= $6 * 6 * 6 * 6 = 6^{4}$

n(S) = $6^{4}$

Let X be the event that all dice show the same face.

X = { (1,1,1,1,), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)}

n(X) = 6

Hence required probability = $\frac{n (X)}{n (S)} = \frac{6}{6^{4}}$ = $\frac{1}{216}$

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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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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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Filed Under: Probability

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