Two slips of papers are drawn from a hat having four slips labelled with $7, 8, 9 a n d 10$ . What is the probability of obtaining $7$ and then $9$ (the first slip is replaced before drawing the second slip) ?

\frac{1}{16}

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\frac{1}{8}

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\frac{1}{4}

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\frac{3}{5}

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Solution

The correct option is A $\frac{1}{16}$
$P r o b a b i l i t y o f o p e n i n g 7 = 1 / 4 P r o b a b i l i t y o f o p e n i n g 9 = 1 / 4 (s i n c e s l i p h a s b e e n r e p l a c e d) P r o b a b i l i t y o f g e t t i n g 7 a n d t h e n 9 = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$

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Two slips of papers are drawn from a hat having four slips labelled with 7,8,9and10. What is the probability of obtaining 7 and then 9 (the first slip is replaced before drawing the second slip) ?

The correct option is A 116Probabilityofopening7=1/4Probabilityofopening9=1/4(sincesliphasbeenreplaced)Probabilityofgetting7andthen9=14×14=116

Two slips of papers are drawn from a hat having four slips labelled with $7, 8, 9 a n d 10$ . What is the probability of obtaining $7$ and then $9$ (the first slip is replaced before drawing the second slip) ?

The correct option is A $\frac{1}{16}$
$P r o b a b i l i t y o f o p e n i n g 7 = 1 / 4 P r o b a b i l i t y o f o p e n i n g 9 = 1 / 4 (s i n c e s l i p h a s b e e n r e p l a c e d) P r o b a b i l i t y o f g e t t i n g 7 a n d t h e n 9 = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$