Jamia Millia Islamia MCA Probability Previous Year Questions (PYQs) – Page 1 of 3
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The probability that at least one of the events A and B occurs is 0.6.
If A and B occur simultaneously with probability 0.2,
then $P(\bar{A}) + P(\bar{B})$ is:
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Solution
$P(A \cup B) = 0.6$, $P(A \cap B) = 0.2$
We know, $P(\bar{A}) + P(\bar{B}) = 2 - P(A) - P(B)$
and $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$\Rightarrow P(A) + P(B) = 0.6 + 0.2 = 0.8$
$\Rightarrow P(\bar{A}) + P(\bar{B}) = 2 - 0.8 = 1.2$
$\boxed{\text{Answer: (C) 1.2}}$
Three persons A, B, and C fire at a target in turn, starting with A.
Their probabilities of hitting the target are $0.4,\ 0.3,\ 0.2$ respectively.
The probability of exactly two hits is:
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Solution
**Solution:**
Let $A,B,C$ denote hitting events.
Probability of exactly 2 hits:
\[
P = P(A,B,\bar{C}) + P(A,\bar{B},C) + P(\bar{A},B,C)
\]
$= (0.4)(0.3)(0.8) + (0.4)(0.7)(0.2) + (0.6)(0.3)(0.2)$
$= 0.096 + 0.056 + 0.036 = 0.188$
$\boxed{\text{Answer: (B) 0.188}}$
A and B are two students. Their chances of solving a problem correctly are
$\dfrac{1}{3}$ and $\dfrac{1}{4}$ respectively.
If the probability of their making a **common error** is $\dfrac{1}{20}$,
and they obtain the same answer, then the probability that their answer is correct is:
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Solution
**Solution:**
Let
$A_1 =$ A correct, $A_2 =$ A wrong
$B_1 =$ B correct, $B_2 =$ B wrong
Then,
$P(A_1) = \dfrac{1}{3}, \quad P(A_2) = \dfrac{2}{3}$
$P(B_1) = \dfrac{1}{4}, \quad P(B_2) = \dfrac{3}{4}$
They give the **same answer** if both are correct or both are wrong.
\[
P(\text{same}) = P(A_1,B_1) + P(A_2,B_2)
\]
Assuming independence except for the given *common error*,
\[
P(A_1,B_1) = \dfrac{1}{3} \cdot \dfrac{1}{4} = \dfrac{1}{12},
\qquad P(A_2,B_2) = \dfrac{1}{20}
\]
Then total
\[
P(\text{same}) = \dfrac{1}{12} + \dfrac{1}{20}
= \dfrac{8}{60} = \dfrac{2}{15}
\]
Hence,
\[
P(\text{correct | same})
= \dfrac{P(A_1,B_1)}{P(\text{same})}
= \dfrac{\tfrac{1}{12}}{\tfrac{2}{15}}
= \dfrac{15}{24} = \dfrac{5}{8}
\]
Let the quadratic equation $ax^2 + bx + c = 0$
where $a, b, c$ are obtained by rolling a dice thrice.
What is the probability that the equation has equal roots?
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Solution
**Solution:**
For equal roots, discriminant $b^2 - 4ac = 0$.
Each of $a, b, c$ can take values $1$ to $6$.
Total outcomes = $6^3 = 216$.
For given $a, c$, $b^2 = 4ac$ must be a perfect square $\le 36$.
Possible $(a, c)$ pairs that make $b^2$ a perfect square:
$(1,1),(1,4),(1,9),(1,16),(1,25),(1,36)$ within dice limit $(1,1)$, $(1,2)$, $(2,1)$, $(3,3)$, $(4,1)$ only valid → 6 cases out of 216.
Hence probability = $\dfrac{6}{216} = \dfrac{1}{36}$.
$\boxed{\text{Answer: (C) }\dfrac{1}{36}}$
A card is drawn from a pack of 52 cards.
A gambler bets that it is a spade or an ace.
What are the odds against his winning this bet?
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Solution
Total cards = 52
Favorable cards = 13 spades + 3 other aces = 16
Odds **against** = $\dfrac{\text{unfavorable}}{\text{favorable}} = \dfrac{52 - 16}{16} = \dfrac{36}{16} = 9:4.$
Two dice are thrown simultaneously.
The probability of obtaining a total score of $5$ is
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Solution
Possible outcomes → $(1,4),(2,3),(3,2),(4,1)$ → $4$ cases.
Total $=36$ outcomes.
Probability $=\dfrac{4}{36}=\dfrac{1}{9}.$
Three of the six vertices of a regular hexagon are chosen at random.
The probability that the triangle formed is equilateral is equal to
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Solution
Total ways $={6 \choose 3}=20$.
Equilateral triangles possible $=2$.
Required probability $=\dfrac{2}{20}=\dfrac{1}{10}.$
A pair of fair dice is thrown independently 3 times. The probability of getting a score of exactly 9 twice is
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Solution
$P(\text{sum} = 9) = \dfrac{4}{36} = \dfrac{1}{9}$
$\Rightarrow P(\text{exactly 2 times}) = \binom{3}{2}\left(\dfrac{1}{9}\right)^2\left(\dfrac{8}{9}\right) = 3 \times \dfrac{1}{81} \times \dfrac{8}{9} = \dfrac{8}{243}$
A black and a red die are rolled together. What is the conditional probability of obtaining the sum $8$, given that the red die resulted in a number less than $4$?
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Solution
Condition: red die ∈ {1,2,3} → sample size $3 \times 6 = 18$ equiprobable outcomes.
Sum $8$ occurs with $(r,b)=(2,6),(3,5)$ → 2 favorable outcomes.
$P(\text{sum }8 \mid r<4) = \dfrac{2}{18} = \dfrac{1}{9}$.
$\boxed{\text{Answer: (C) } \tfrac{1}{9}}$
Three houses are available in a locality. Three persons apply for the houses.
Each applies for one house without consulting others.
The probability that all the three apply for the same house is...
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Solution
Each person can choose any of 3 houses ⇒ total cases $= 3^3 = 27.$
All three apply for the same house ⇒ favorable cases $= 3.$
So $P = \dfrac{3}{27} = \dfrac{1}{9}.$
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