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Probability

Three numbers...

Question

Three numbers are chosen at random withput replacement from ${1, 2, . . ., 15}$ . Let $E_{1}$ be the event that minimum of the chosen numbers is 5 and $E_{2}$ their maximum is 10 then

P (E_{1}) = \frac{9}{91}

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P (E_{2}) = \frac{36}{455}

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P (E_{1} \cap E_{2}) = \frac{4}{455}

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P (E_{1} | E_{2}) = \frac{1}{9}

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Solution

The correct options are
A $P (E_{1} \cap E_{2}) = \frac{4}{455}$
B $P (E_{1} | E_{2}) = \frac{1}{9}$
C $P (E_{2}) = \frac{36}{455}$
D $P (E_{1}) = \frac{9}{91}$
$P (E_{1}) = P$ (choosing 5 and two tickets from 6 to 15)
$= \frac{^{10} C_{2}}{^{15} C_{3}} = \frac{10 \times 9}{2} \times \frac{3 \times 2}{15 \times 14 \times 13} = \frac{9}{91}$
$P (E_{2}) = P$ (choosing 10 and two tickets from 1 to 9)
$= \frac{^{9} C_{2}}{^{15} C_{3}} = \frac{9 \times 8}{2} \times \frac{3 \times 2}{15 \times 14 \times 13} = \frac{36}{455}$
$P (E_{1} \cap E_{2}) = P$ (choosing 5 to 10 and one ticket from 6 to 9)
$= \frac{^{4} C_{1}}{^{15} C_{3}} = 4 \times \frac{3 \times 2}{15 \times 14 \times 13} = \frac{4}{455}$
$P (E_{1} | E_{2}) = \frac{P (E_{1} \cap E_{2})}{P (E_{2})} = \frac{1}{9}$

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Similar questions

Three numbers are chosen at random without replacement from {1, 2, ......, 15}. Let $E_{1}$ be the event that minimum of the chosen numbers is 5 and $E_{2}$ be that their maximum is 10 then

E_{1}

and

E_{2}

are two events such that

P (E_{1}) = \frac{1}{4}

P (\frac{E_{2}}{E_{1}}) = \frac{1}{2}

and

P (\frac{E_{1}}{E_{2}}) = \frac{1}{4}

, then

Q. If

E_{1}

and

E_{2}

are two events such that

P (E_{1}) = \frac{1}{4}, P (\frac{E_{2}}{E_{1}}) = \frac{1}{2}

and

P (\frac{E_{1}}{E_{2}}) = \frac{1}{4},

then which of the following is/are correct?

E_{1} a n d E_{2}

are events in a probability space satisfying the following constraints

P (E_{1}) = P (E_{2}); P (E_{1} \cup E_{2}) = 1; E_{1} a n d E_{2}

are independent then

P (E_{1}) =

Q. If

E_{1}

and

E_{2}

are two events such that

P (E_{1}) = 1 / 4, P (E_{2} / E_{1}) = 1 / 2

and

P (E_{1} / E_{2}) = 1 / 4

, then

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Three numbers are chosen at random withput replacement from {1,2,...,15}. Let E1 be the event that minimum of the chosen numbers is 5 and E2 their maximum is 10 then

Three numbers are chosen at random withput replacement from ${1, 2, . . ., 15}$ . Let $E_{1}$ be the event that minimum of the chosen numbers is 5 and $E_{2}$ their maximum is 10 then