$$6$$ ordinary dice are rolled. The probability that at least half of them will show at least $$3$$ is :
A.
$$41 \times \frac{{{2^4}}}{{{3^6}}}$$
B.
$$\frac{{{2^4}}}{{{3^6}}}$$
C.
$$20 \times \frac{{{2^4}}}{{{3^6}}}$$
D.
none of these
Answer :
$$41 \times \frac{{{2^4}}}{{{3^6}}}$$
Solution :
The probability of getting at least $$3$$ in a throw $$ = \frac{4}{6} = \frac{2}{3}$$
$$\therefore $$ the required probability
$$ = {}^6{C_3}.{\left( {\frac{2}{3}} \right)^3}.{\left( {\frac{1}{3}} \right)^3} + {}^6{C_4}.{\left( {\frac{2}{3}} \right)^4}.{\left( {\frac{1}{3}} \right)^2} + {}^6{C_5}.{\left( {\frac{2}{3}} \right)^5}.\frac{1}{3} + {}^6{C_6}.{\left( {\frac{2}{3}} \right)^6}$$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$