JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Probability XII

Let

E

and

F

be two independent events. The probability that exactly one of them occurs is

\frac{11}{25}

and the probability of none of them occurring is

\frac{2}{25}

. If

P (T)

denotes the probability of occurrence of the event

T

, then

Answer: (A), (D)

Solution

Let

P (E) = e

and

P (F) = f

.

Probability that exactly one occurs:

P (E) P (\overset{ˉ}{F}) + P (\overset{ˉ}{E}) P (F) = e (1 - f) + (1 - e) f = e + f - 2 e f = \frac{11}{25}

... (1)

Probability that neither occurs:

P (\overset{ˉ}{E}) P (\overset{ˉ}{F}) = (1 - e) (1 - f) = 1 - e - f + e f = \frac{2}{25}

... (2)

From (2):

e + f - e f = 1 - \frac{2}{25} = \frac{23}{25}

... (3)

From (1):

e + f - 2 e f = \frac{11}{25}

... (4)

Subtracting (4) from (3):

e f = \frac{12}{25}

From (3):

e + f = \frac{23}{25} + \frac{12}{25} = \frac{35}{25} = \frac{7}{5}

e

and

f

are roots of

t^{2} - \frac{7}{5} t + \frac{12}{25} = 0

, or

25 t^{2} - 35 t + 12 = 0

.

Factoring:

(5 t - 3) (5 t - 4) = 0

, giving

t = \frac{3}{5}

t = \frac{4}{5}

.

Therefore,

(P (E), P (F)) = (\frac{4}{5}, \frac{3}{5})

(\frac{3}{5}, \frac{4}{5})

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · One or more correct options