Test Index

BITSAT 2014 Paper

Section: Mathematics

Question : 119 of 150

Marks: +1, -0

\frac{7}{12}

, then the value of n is.

Solution:

Let

E_1

denote the event "a coin with head on both sides is selected" and

E_2

denotes the event " a fair coin is selected". Let A be the event " he toss, results in heads".

∴

P(E_1)

\frac{1}{n+1}

P(E_2)

\frac{n}{n+1}

and

P(A/E_1)

= 1 ,

P(A/E_2)

\frac{1}{2}

∴ P (A) =

P(E_1)

(A/E_1)

P(E_2) (A/E_2)

⇒

\frac{7}{12}

\frac{1}{n+1}

× 1 +

\frac{n}{n+1}

\frac{1}{2}

⇒ 14n + 14 = 24 + 12n ⇒ n = 5

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