CUET UG Mathematics Applied Probability Previous Year Questions (PYQs)
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Showing 7 questions
Two dice are thrown simultaneously. If X denotes the number of fours, then the expectation of X will be:
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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 :
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Solution
For two events $A, B$
$P(A\cup B)=\frac{7}{12}, ; P(A)=\frac{5}{12}, ; P(B)=\frac{3}{12}$
Then $P(A\cap B)=$
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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:
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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:
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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
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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?
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Solution
CUET UG Mathematics Applied
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CUET UG Mathematics Applied
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
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