All the spades are taken out from a pack of cards. From these cards, cards are drawn one by one without replacement till the ace of spades comes. The probability that the ace comes in the $${4^{th}}$$ draw is :
A.
$$\frac{1}{{13}}$$
B.
$$\frac{{12}}{{13}}$$
C.
$$\frac{4}{{13}}$$
D.
none of these
Answer :
$$\frac{1}{{13}}$$
Solution :
The probability of not drawing the ace in the first draw, in the second draw and in the third draw are $$\frac{{12}}{{13}},\,\frac{{11}}{{12}}$$ and $$\frac{{10}}{{11}}$$ respectively.
The probability of drawing ace of spades in the fourth draw $$ = \frac{1}{{10}}$$
$$\therefore $$ the required probability $$ = \frac{{12}}{{13}} \times \frac{{11}}{{12}} \times \frac{{10}}{{11}} \times \frac{1}{{10}} = \frac{1}{{13}}$$
Releted MCQ Question on Statistics and Probability >> Probability
Releted Question 1
Two fair dice are tossed. Let $$x$$ be the event that the first die shows an even number and $$y$$ be the event that the second die shows an odd number. The two events $$x$$ and $$y$$ are:
Two events $$A$$ and $$B$$ have probabilities 0.25 and 0.50 respectively. The probability that both $$A$$ and $$B$$ occur simultaneously is 0.14. Then the probability that neither $$A$$ nor $$B$$ occurs is
The probability that an event $$A$$ happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event $$A$$ happens at least once is
If $$A$$ and $$B$$ are two events such that $$P(A) > 0,$$ and $$P\left( B \right) \ne 1,$$ then $$P\left( {\frac{{\overline A }}{{\overline B }}} \right)$$ is equal to
(Here $$\overline A$$ and $$\overline B$$ are complements of $$A$$ and $$B$$ respectively).
A.
$$1 - P\left( {\frac{A}{B}} \right)$$
B.
$$1 - P\left( {\frac{{\overline A }}{B}} \right)$$
C.
$$\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}}$$
D.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$