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Q:

In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

A) 2/7	B) 5/7
C) 1/5	D) 1/2

Answer & Explanation Answer: A) 2/7

Explanation:

Total number of outcomes possible, n(S) = 10 + 25 = 35

Total number of prizes, n(E) = 10

$P (E) = \frac{n (E)}{n (S)} = \frac{10}{35} = \frac{2}{7}$

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Q:

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

A) 21/46	B) 1/5
C) 3/25	D) 1/50

Answer & Explanation Answer: A) 21/46

Explanation:

Let , S - sample space E - event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25

= $25 C_{3}$

= 2300.

n(E) = $10 C_{1} \times 15 C_{2}$ = 1050.

$\therefore$ P(E) = n(E)/n(s) = 1050/2300 = 21/46

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Q:

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

A) 21/46	B) 1/5
C) 3/25	D) 1/50

Answer & Explanation Answer: A) 21/46

Explanation:

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25

= $25 C_{3}$ = 2300.

n(E)= $10 C_{1} * 15 C_{2}$ = 1050.

P(E) = n(E)/n(s) = 1050/2300 = 21/46

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Q:

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

A) 3/4	B) 3/8
C) 5/16	D) 2/7

Answer & Explanation Answer: A) 3/4

Explanation:

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

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

n(E) = 27.

P(E) = n(E)/n(S) = 27/36 = 3/4.

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Q:

Three unbiased coins are tossed. What is the probability of getting at most two heads?

A) 3/4	B) 7/8
C) 1/2	D) 1/4

Answer & Explanation Answer: B) 7/8

Explanation:

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

P(E) =n(E)/n(S)=7/8.

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Q:

What would be the least number of years in which the simple interest on Rs.2600 at % will be an exact number of rupees ?

A) 2	B) 3
C) 4	D) 5

Answer & Explanation Answer: B) 3

Explanation:

$S . I = R s . [2600 * \frac{20}{3} * \frac{1}{100} * t]$

$S . I = R s . [\frac{520}{3} * t]$

t=3

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Filed Under: Compound Interest
Exam Prep: Bank Exams
Job Role: Bank PO

Q:

What is the probability of getting a sum 9 from two throws of a dice?

A) 1/2	B) 3/4
C) 1/9	D) 2/9

Answer & Explanation Answer: C) 1/9

Explanation:

In two throws of a die, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

P(E) =n(E)/n(S)=4/36=1/9.

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Q:

A sum was put at simple interest at a certain rate for 10 years . Had it been put at 5% higher rate , it would have fetched Rs.600 more. What was the Sum?

A) Rs.1200	B) Rs.1300
C) Rs.1400	D) Rs.1500

Answer & Explanation Answer: A) Rs.1200

Explanation:

At 5% more rate, the increase in S.I for 10 years = Rs.600 (given)

So, at 5% more rate, the increase in SI for 1 year = 600/10 = Rs.60/-

i.e. Rs.60 is 5% of the invested sum

So, 1% of the invested sum = 60/5

Therefore, the invested sum = 60 × 100/5 = Rs.1200

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Filed Under: Simple Interest
Exam Prep: Bank Exams
Job Role: Bank PO

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