Jamia Millia Islamia Probability Previous Year Questions (PYQs) – Page 2 of 3
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In a unique hockey series between India & Pakistan, they decide to play on till a team wins 5 matches.
The number of ways in which the series can be won if no match ends in a draw is:
The number of ways in which the series can be won if no match ends in a draw is:
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Solution
To win, one team must reach 5 wins before the other.
Total possible ways =
$2 \times \sum_{k=0}^{4} \binom{4+k}{k} = 2(1 + 5 + 15 + 35 + 70) = 252$
Let two fair six-faced dice $A$ and $B$ be thrown simultaneously.
Let $E_1$ be the event that die $A$ shows 4,
$E_2$ the event that die $B$ shows 2,
and $E_3$ the event that the sum of the two numbers on the dice is odd.
Which statement is false?
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Solution
$P(E_1) = P(E_2) = \dfrac{1}{6}, \quad P(E_3) = \dfrac{1}{2}$
$P(E_1 \cap E_2 \cap E_3) = 0$ (since $4 + 2 = 6$, even)
$P(E_1)P(E_2)P(E_3) = \dfrac{1}{72}$
So they are pairwise independent, but not mutually independent.
Two dice are thrown. The probability that the sum of the numbers on two dice is 7 is …
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Solution
Total outcomes = $6 \times 6 = 36$.
Favorable outcomes for sum = 7 are $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ → 6 outcomes.
Probability $= \dfrac{6}{36} = \dfrac{1}{6}$.
A single letter is selected at random from the word “JAMIA”. The probability that it is a vowel is …
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Solution
Letters in “JAMIA” = J, A, M, I, A → total 5 letters.
Vowels = A, I, A → 3 vowels.
Probability $= \dfrac{3}{5}$.
One die and a coin are tossed simultaneously. The probability of getting 6 on die and head on coin is …
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Solution
Probability of 6 on die $= \dfrac{1}{6}$,
Probability of head on coin $= \dfrac{1}{2}$.
Required probability $= \dfrac{1}{6} \times \dfrac{1}{2} = \dfrac{1}{12}$.
An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement.
What is the probability that both drawn balls are black?
What is the probability that both drawn balls are black?
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Solution
Total balls $= 15$
Probability of drawing two black balls:
$P = \dfrac{10}{15} \times \dfrac{9}{14} = \dfrac{90}{210} = \dfrac{3}{7}$
A box has 5 black and 3 green shirts. One shirt is picked randomly and put in another box.
The second box has 3 black and 5 green shirts. Now a shirt is picked from the second box.
What is the probability of it being a black shirt?
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Solution
29/72
If two fair dice are rolled, what is the probability of getting a sum of 7?
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Solution
Total possible outcomes:
6× 6 = 36
Favourable outcomes giving sum 7:
(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)
Total favourable outcomes = 6
Probability:
P = 6/36 = 1/6
If $A$ and $B$ are independent events, then $P(A \cap B)$ is:
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Solution
$P(A \cap B) = P(A)P(B)$
(b) Sample space
$S = {HH, HT, TH, TT}$
Favourable outcomes
$P = \frac{3}{4}$
The probability of getting at least one head when tossing two fair coins is:
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Solution
E(X)=5×0.3
E(X)=1.5
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