CUET UG Mathematics Applied Probability Previous Year Questions (PYQs)
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
Two dice are thrown simultaneously. If X denotes the number of fours, then the expectation of X will be:
1
2
3
4
Solution
If the random variable $X$ has the following distribution :
| X | P(X) |
|---|---|
| 0 | k |
| 1 | 2k |
| 2 | 3k |
| otherwise | 0 |
Match List-I with List-II :
| List-I | List-II |
|---|---|
| (A) $k$ | (I) $\dfrac{5}{6}$ |
| (B) $P(X<2)$ | (II) $\dfrac{4}{3}$ |
| (C) $E(X)$ | (III) $\dfrac{1}{2}$ |
| (D) $P(1\le X\le 2)$ | (IV) $\dfrac{1}{6}$ |
Choose the correct answer from the options given below :
1
2
3
4
Solution
For two events $A, B$
$P(A\cup B)=\frac{7}{12}, ; P(A)=\frac{5}{12}, ; P(B)=\frac{3}{12}$
Then $P(A\cap B)=$
1
2
3
4
Solution
We know,
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
Substitute values:
$\frac{7}{12}=\frac{5}{12}+\frac{3}{12}-P(A\cap B)$
$\frac{7}{12}=\frac{8}{12}-P(A\cap B)$
So,
$P(A\cap B)=\frac{8}{12}-\frac{7}{12}$
$=\frac{1}{12}$
In a game, a child will win Rs $5$ if he gets all heads or all tails when three coins are tossed simultaneously and he will lose Rs $3$ for all other cases.
The expected amount to lose in the game is:
1
2
3
4
Solution
Total outcomes when 3 coins tossed:
$2^3=8$
Favourable (all heads or all tails):
HHH, TTT → 2 outcomes
So,
$P(\text{win})=\frac{2}{8}=\frac{1}{4}$
$P(\text{lose})=\frac{6}{8}=\frac{3}{4}$
Expected value:
$E=5\left(\frac{1}{4}\right)+(-3)\left(\frac{3}{4}\right)$
$=\frac{5}{4}-\frac{9}{4}$
$=-\frac{4}{4}$
$=-1$
So expected loss = Rs. 1
If events $A$ and $B$ are independent, then identify the correct statements
(A) $A$ and $B$ must be mutually exclusive
(B) The sum of their probabilities must be equal to $1$
(C) $P(A)\cdot P(B) = P(A\cap B)$
(D) $A'$ and $B'$ are also independent
Choose the correct answer from the options given below:
1
2
3
4
Solution
For independent events
$P(A\cap B)=P(A)P(B)$
So statement (C) is **true**.
Independent events need not be mutually exclusive.
Therefore (A) is **false**.
Also $P(A)+P(B)$ need not be $1$.
Hence (B) is **false**.
If $A$ and $B$ are independent then their complements $A'$ and $B'$ are also independent.
Hence (D) is **true**.
Correct statements: **C and D**
If $A$ and $B$ are two independent events with
$P(A)=\frac{3}{5}$ and $P(B)=\frac{4}{9}$, then
$P(A' \cap B')$ is equal to
1
2
3
4
Solution
Since $A$ and $B$ are independent,
$P(A\cap B)=P(A)P(B)$
Now
$P(A') = 1 - P(A)$
$P(B') = 1 - P(B)$
So
$P(A') = 1-\frac{3}{5}=\frac{2}{5}$
$P(B') = 1-\frac{4}{9}=\frac{5}{9}$
For independent events, complements are also independent.
Hence
$P(A'\cap B') = P(A')P(B')$
$=\frac{2}{5}\times\frac{5}{9}$
$=\frac{2}{9}$
A die is rolled thrice. What is the probability of getting a number greater than 4
in the first and second throw of the dice, and a number less than 4 in the third throw?
1
2
3
4
Solution
CUET UG Mathematics Applied
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
CUET UG Mathematics Applied
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
