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IBPS PO Mains 2018 Paper for online practice

DIRECTIONS (Qs.98-100):

There are three quantities provided in the questions. You have to find out the values of the quantities and compare them according to the given codes as follows

@ ® >

& ⇒ <

* ⇒ >

\backslash

\Rightarrow < \\ \# \Rightarrow = \text{(or relationship can't be established)} \\ \text{Example : } \\ \text{Quantity I: }23 + 33 + 43 \\ \text{Quantity II: }62 \times 22 \\ \text{Quantity III: }63 - 23 \times 32 \\ \text{(a) @, }

\ \text{(b) \&, \#} \ \text{ (c) }

, #

(d) &, *

(e) @, #

Quantity I < Quantity II = Quantity III

So, answer is (b)

Question : 100 of 155

Marks: +1, -0

Quantity I – In a question paper probability of a question not being answered by three students is 0.5, 0.4, 0.7 respectively. Find the probability that at most two students will solve the question.

Quantity II – Dipak have 5 black balls & 7 white balls, in his bag. If three balls are drawn at random from bag, then find probability of getting at least 1 black ball.

Quantity III – Nirmal speaks the truth 4 out of 5 times, and Puja speaks the truth 6 out of 7 times. What is the probability that they will contradict each other in stating the same fact?

(@, &)
(#, @)
(@, @)
(&, a)
(*, #)

Validate

Solution:

Quantity I –

Probability of at most two students will solve the question

= 1 – probability of all three students will solve the question

= 1 – (0.5) × (0.6) × (0.3)

= 1 – 0.09

= 0.91

Quantity II –

Total balls = 5 + 7 = 12

Probability of getting at least 1 black ball = 1 - probability of no black ball.

1-\frac{7}{44}

\frac{37}{44}

~ 0.84

Quantity III –

P (Nirmal speak truth) =

\frac{4}{5}

P (Puja speak truth) =

\frac{6}{7}

Required probability

\frac{4}{5} \times \frac{1}{7} + \frac{1}{5} \times \frac{6}{7}

\frac{10}{35}

\frac{2}{7}

~ 0.28

Quantity I > Quantity II > Quantity III

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