India plays two matches each with West Indies and Australia. In any match the probabilities of India getting, points 0, 1 and 2 are 0.45, 0.05 and 0.50 respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is
A.
0.8750
B.
0.0875
C.
0.0625
D.
0.0250
Answer :
0.0875
Solution :
$$P$$ (at least 7 pts) = $$P$$ (7pts) + $$P$$ (8 pts)
[ $$\because $$ At most 8 pts can be scored. ]
Now 7 pts can be scored by scoring 2 pts in 3 matches and 1 pt. in one match, similarly 8 pts can be scored by scoring 2 pts in each of the 4 matches.
$$\eqalign{
& \therefore \,\,{\text{Req}}{\text{. prob}}{\text{.}} = {\,^4}{C_1} \times {\left[ {P\left( {2{\text{pts}}} \right)} \right]^3}P\left( {1{\text{pts}}} \right) + {\left[ {P\left( {2{\text{pts}}} \right)} \right]^4} \cr
& = 4{\left( {0.5} \right)^3} \times 0.05 + {\left( {0.50} \right)^4} \cr
& = 0.0250 + 0.0625 \cr
& = 0.0875 \cr} $$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$