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BITSAT 2015 Paper

Section: Mathematics

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Question : 88 of 150

Marks: +1, -0

A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides, whereas the remaining (n + 1) coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is 31/42, then n is equal to

10
11
12
13

Solution:

Total number of coins= 2n+1

Consider the following events:

E_1

= Getting a coin having head on both sides from the bag.

E_2

= Getting a fair coin from the bag

A = Toss results in a head

Given: P (A) =

\frac{31}{42}

, P

(E_1)

=

\frac{n}{2n+1}

and

P(E_2)

=

\frac{n+1}{2n+1}

Then,

P (A) = P

(E_1)

P (A/

E_1

) + P

(E_2)

P (A/

E_2

)

⇒

\frac{31}{42}

=

\frac{n}{2n+1}

× 1 +

\frac{n+1}{2n+1} \times \frac{1}{2}

\frac{31}{42}

=

\frac{n}{2n+1}

+

\frac{n+1}{2(2n+1)}

⇒

\frac{31}{42}

=

\frac{3n+1}{2(2n+1)}

⇒

\frac{31}{21}

=

\frac{3n+1}{2n+1}

n = 10

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