A flashlight has $8$ batteries out of which $3$ are dead. If two batteries are selected without replacement and tested, the probability that both are dead is (A) $\dfrac{33}{56}$ (B) $\dfrac{9}{64}$ (C) $\dfrac{1}{14}$ (D) $\dfrac{3}{28}$

> [!TLDR] > This is a hypergeometric probability problem — selecting without replacement from a finite set with two types (dead/working). The probability that both selected batteries are dead is $\frac{3}{28}$, which corresponds to option **(D)**. The key here is recognising that we are drawing **without replacement** from a small population. When you pick the first battery, the total number of batteries and the number of dead ones both decrease, so the second draw's probability depends on the first. This is not a binomial situation (where draws are independent) — it's a **hypergeometric** situation. Let’s walk through it. 1. **Total batteries and dead ones** We have $8$ batteries total, of which $3$ are dead. So $5$ are working. 2. **Probability that the first battery is dead** On the first pick, there are $3$ dead out of $8$ total. So $$P(\text{first dead}) = \frac{3}{8}$$ 3. **Probability that the second battery is dead, given the first was dead** After removing one dead battery, we have $2$ dead left and only $7$ batteries remaining. So $$P(\text{second dead} \mid \text{first dead}) = \frac{2}{7}$$ 4. **Multiply for the joint probability** Since we want both to be dead, we multiply the conditional probabilities: $$P(\text{both dead}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}$$ 5. **Simplify the fraction** $\frac{6}{56}$ simplifies by dividing numerator and denominator by $2$: $$\frac{6}{56} = \frac{3}{28}$$ …

A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and te

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Concept understanding — Hypergeometric Probability

Hypergeometric Probability: Drawing Without Replacement

You have a bag of 20 marbles: 12 red and 8 blue. You pick 5 without putting any back. What's the chance exactly 3 are red?

This is what the hypergeometric distribution handles. Unlike the binomial distribution, the trials are not independent — each draw changes the composition of the bag.

The Intuition

When you draw without replacement, the probability of a red on the second draw depends on the first: take a red first and fewer reds remain, so the next red is less likely. Hypergeometric probability captures exactly this dependency.

The setup: "I have a finite population split into two groups. I take a sample without replacement. What's the probability my sample has exactly $k$ items from the first group?"

The Precise Statement

P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}

Where:

$N$ = total items in the population (20 marbles)
$K$ = number of "success" items ( $12$ red)
$n$ = number drawn (sample size, $5$ )
$k$ = successes wanted in the sample ( $3$ reds)

Why This Formula Makes Sense

The denominator $\binom{N}{n}$ counts all ways to choose $n$ items from $N$ — the equally likely outcomes. The numerator counts favourable ones:

$\binom{K}{k}$ : choose $k$ reds from the $K$ reds
$\binom{N-K}{n-k}$ : choose the remaining $n-k$ from the $N-K$ blues

Multiplying pairs each way of picking reds with each way of picking blues.

Worked Example

$N=20$ , $K=12$ , $n=5$ , $k=3$ :

P(\text{exactly 3 reds}) = \frac{\binom{12}{3} \binom{8}{2}}{\binom{20}{5}} = \frac{220 \times 28}{15504} = \frac{6160}{15504} \approx 0.397

About 39.7%.

Watch out

A common mistake is using the binomial formula here. Binomial assumes independent trials (drawing with replacement). With $p = 12/20 = 0.6$ it gives $\binom{5}{3}(0.6)^3(0.4)^2 \approx 0.346$ — close but wrong. The gap grows as the sample becomes a larger fraction of the population.

When to Use Hypergeometric …

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This is a hypergeometric probability problem — selecting without replacement from a finite set with two types (dead/working). The probability that both selected batteries are dead is $\frac{3}{28}$ , which corresponds to option (D).

The key here is recognising that we are drawing without replacement from a small population. When you pick the first battery, the total number of batteries and the number of dead ones both decrease, so the second draw's probability depends on the first. This is not a binomial situation (where draws are independent) — it's a hypergeometric situation.

Let’s walk through it.

Total batteries and dead ones

We have $8$ batteries total, of which $3$ are dead. So $5$ are working.
Probability that the first battery is dead

On the first pick, there are $3$ dead out of $8$ total. So

P(\text{first dead}) = \frac{3}{8}

Probability that the second battery is dead, given the first was dead After removing one dead battery, we have $2$ dead left and only $7$ batteries remaining. So

P(\text{second dead} \mid \text{first dead}) = \frac{2}{7}

Multiply for the joint probability Since we want both to be dead, we multiply the conditional probabilities:

P(\text{both dead}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}

Simplify the fraction $\frac{6}{56}$ simplifies by dividing numerator and denominator by $2$ : $\frac{6}{56} = \frac{3}{28}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
$P$
›Reveal solutionSolution
This tests conditioning on which bag contributes the "extra" red or black ball, using combinations for the two-ball draw from bag $Q$ . The probability of exactly two black and one red overall is $\dfrac{25}{54}$ .

Concept and Intuition

The three balls come from two different draws (1 from $P$ , 2 from $Q$ ), so split into mutually exclusive cases based on the colour drawn from $P$ , since that determines what's needed from $Q$ to reach "two black, one red" overall.

Step-by-Step Solution

Bag $P$ : 4 red, 5 black (9 total). Bag $Q$ : 3 red, 6 black (9 total).

Case 1 — ball from $P$ is red (prob $\tfrac49$ ): then both balls from $Q$ must be black to get "2 black, 1 red" overall. $P(\text{2 black from }Q)=\dfrac{\binom62}{\binom92}=\dfrac{15}{36}=\dfrac5{12}$ . \quad Case 1 probability $=\tfrac49\times\tfrac5{12}=\tfrac{20}{108}=\tfrac5{27}$ .

Case 2 — ball from $P$ is black (prob $\tfrac59$ ): then $Q$ must contribute exactly 1 black and 1 red. $P(\text{1 black,1 red from }Q)=\dfrac{\binom61\binom31}{\binom92}=\dfrac{18}{36}=\tfrac12$ . \quad Case 2 probability $=\tfrac59\times\tfrac12=\tfrac5{18}$ . …
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AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
$\dfrac{343}{1720}$
›Reveal solutionSolution
This tests probability of drawing all-same-color balls without replacement, computed via combinations. The probability is $\dfrac{7}{44}$ .

Concept and Intuition

When drawing several balls without replacement (and order of drawing doesn't affect which SET of balls you end up with), the cleanest approach is combinations: the probability that all 3 drawn balls are green is the number of ways to choose 3 green balls out of the 7 available, divided by the number of ways to choose any 3 balls out of all 12.

Step-by-Step Solution

Total balls $=7$ green $+5$ black $=12$ .

Ways to choose 3 balls from 12 (total sample space): $\binom{12}{3}=\dfrac{12\times11\times10}{3\times2\times1}=220$ .

Ways to choose 3 green balls from the 7 green ones: $\binom{7}{3}=\dfrac{7\times6\times5}{3\times2\times1}=35$ .

$P(\text{all 3 green}) = \dfrac{35}{220}$ .

Simplify: $\dfrac{35}{220}=\dfrac{7}{44}$ (dividing by 5).

Common Mistakes …
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AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$\dfrac{34}{55}$
›Reveal solutionSolution
"At most one rotten" means 0 or 1 rotten among the 3 drawn; combining both cases over $\binom{12}{3}$ total ways gives $\frac{48}{55}$ .

Concept and Intuition

With simultaneous (unordered) selection, we use combinations. "At most one rotten" is the union of two mutually exclusive events: exactly 0 rotten, or exactly 1 rotten. We compute each via the hypergeometric-style counting (choose rotten apples from the 3 rotten, good apples from the 9 good) and add.

Step-by-Step Solution

Basket: 12 apples, 3 rotten, 9 good. Draw 3 simultaneously: total ways $=\binom{12}{3}=220$ .

0 rotten (all 3 good): $\binom{9}{3} = 84$ ways.

1 rotten, 2 good: $\binom{3}{1}\binom{9}{2} = 3 \times 36 = 108$ ways. …
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AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
$\dfrac{20}{27}$
›Reveal solutionSolution
Splitting into the two mutually-exclusive ways to get "2 black, 1 red" across the two bags gives total probability $\dfrac{25}{54}$ .

Concept and Intuition

The overall red/black count depends on which bag contributed which colours, so we enumerate the (bag1-draw, bag2-draw) combinations that together yield exactly 2 black and 1 red, and add their probabilities (law of total probability over independent draws).

Step-by-Step Solution

Bag1: 4 red, 5 black (9 total). Bag2: 3 red, 6 black (9 total), draw 2.

Case 1: bag1 gives red (prob $\frac49$ ), bag2 gives 2 black (prob $\frac{\binom62}{\binom92}=\frac{15}{36}=\frac{5}{12}$ ) → total: 1 red (bag1) + 2 black (bag2) ✓. Probability $=\frac49\times\frac{5}{12}=\frac{20}{108}$ . …
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AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
$\dfrac{1}{5}$
›Reveal solutionSolution
Multiplying successive conditional probabilities of drawing red (without replacement) gives $\dfrac{6}{10}\cdot\dfrac{5}{9}\cdot\dfrac{4}{8}=\dfrac16$ .

Concept and Intuition

For sampling without replacement, the probability of a specific sequence of outcomes is the product of conditional probabilities at each draw, since the pool shrinks and its composition changes after each draw.

Step-by-Step Solution

First ball red: $\dfrac{6}{10}$ .

Second ball red (now 5 red, 9 total left): $\dfrac{5}{9}$ .

Third ball red (now 4 red, 8 total left): $\dfrac{4}{8}$ . …
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AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$V_1$
›Reveal solutionSolution
Tests the complement rule for "not all same category" using combinations; answer is $261/286$ .

Concept and Intuition

It is easier to find the probability of the complementary event (all three bottles from the same variety) and subtract from 1, rather than directly enumerating the many ways to get a mixed selection.

Step-by-Step Solution

Total bottles $=6+3+4=13$ . Total ways to choose 3: $\binom{13}{3}=286$ .

Ways all 3 are $V_1$ : $\binom63=20$ . All $V_2$ : $\binom33=1$ . All $V_3$ : $\binom43=4$ .

$P(\text{all same}) = \dfrac{20+1+4}{286}=\dfrac{25}{286}$ .

$P(\text{not all same}) = 1-\dfrac{25}{286}=\dfrac{286-25}{286}=\dfrac{261}{286}$ . …
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AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
$\dfrac{45}{28}$
›Reveal solutionSolution
Drawing 3 balls without replacement from an urn of 3 black + 5 red gives a hypergeometric distribution for the number of red balls, whose mean is simply $n\cdot K/N = 3\times5/8=15/8$ .

Concept and Intuition

When balls are drawn without replacement, the count of a particular colour follows the hypergeometric distribution, whose mean has the same clean form as the binomial's ( $np$ ), but with $p$ replaced by the population proportion $K/N$ : $E[X] = n\cdot\dfrac{K}{N}$ .

Step-by-Step Solution

Total balls $N=3+5=8$ ; red balls $K=5$ ; balls drawn $n=3$ .

By the hypergeometric mean formula: $E[X] = n\cdot\dfrac{K}{N} = 3\times\dfrac58 = \dfrac{15}{8}$ .

(Cross-check via indicator variables: let $X_i=1$ if the $i$ -th ball drawn is red. Even without replacement, by symmetry $P(X_i=1)=\dfrac58$ for every $i=1,2,3$ individually. So $E[X]=E[X_1]+E[X_2]+E[X_3]=3\times\dfrac58=\dfrac{15}8$ , confirming the formula.)

Common Mistakes …
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AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
$\dfrac{{}^7C_3}{{}^{12}C_9}$
›Reveal solutionSolution
This is a hypergeometric probability: draw 9 balls together from 12 (5 white + 7 black); we want exactly 3 white and 6 black.

Concept and Intuition

When items are drawn simultaneously (not one after another), order doesn't matter, so we count using combinations rather than a sequence of conditional probabilities. The favourable outcomes are: choose 3 white balls out of the 5 available, AND choose 6 black balls out of the 7 available. The total outcomes are all the ways of choosing any 9 balls out of the 12 in the bag.

Step-by-Step Solution

Total ways to draw 9 balls out of 12: ${}^{12}C_9 = {}^{12}C_3 = \dfrac{12\cdot 11\cdot 10}{3\cdot 2\cdot 1} = 220$ .

Favourable ways: choose 3 white out of 5, and 6 black out of 7 (these are independent choices combined by multiplication): ${}^5C_3 = 10, \qquad {}^7C_6 = 7$ Favourable $= 10 \times 7 = 70$ .

Required probability: …
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AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
$\frac{1}{15}$
›Reveal solutionSolution
This is a straightforward combinations-based probability: choose 2 kings out of 4, divided by choosing any 2 cards out of 52.

Concept and Intuition

When drawing cards "at random" without replacement, the probability of an event is (favourable ways to choose the required cards) / (total ways to choose that many cards from the deck), using combinations since order doesn't matter.

Step-by-Step Solution

Total ways to draw 2 cards from 52: $\binom{52}{2}=\dfrac{52\times51}{2}=1326$ .

Favourable ways (both kings) from the 4 kings: $\binom{4}{2}=6$ .

$P(\text{both kings})=\dfrac{6}{1326}=\dfrac{1}{221}$ .

Common Mistakes …
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AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$1$
›Reveal solutionSolution
Drawing 2 balls from 3 white + 3 black, the mean number of white balls is $1$ — exactly half of what's drawn, by symmetry.

Concept and Intuition

By symmetry, since white and black balls are equal in number (3 each), a randomly drawn pair should on average contain equally many white and black balls, so $E[X]=1$ out of the 2 drawn. This can be confirmed rigorously via the hypergeometric distribution formula.

Step-by-Step Solution

Total balls $N=6$ (3 white, 3 black); we draw $n=2$ without replacement. $X=$ number of white balls drawn, so $X\in\{0,1,2\}$ .

$P(X=0)=\dfrac{\binom{3}{0}\binom{3}{2}}{\binom{6}{2}}=\dfrac{1\times3}{15}=\dfrac{3}{15}=\dfrac15$ .

$P(X=1)=\dfrac{\binom{3}{1}\binom{3}{1}}{\binom{6}{2}}=\dfrac{3\times3}{15}=\dfrac{9}{15}=\dfrac35$ .

$P(X=2)=\dfrac{\binom{3}{2}\binom{3}{0}}{\binom{6}{2}}=\dfrac{3\times1}{15}=\dfrac{3}{15}=\dfrac15$ . …
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AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$\dfrac{13}{270725}$
›Reveal solutionSolution
This tests probability of drawing all cards of the same suit, using combinations and one suit choice out of four.

Concept and Intuition

Drawing 4 cards all of the same suit means: pick a suit (4 ways) and then choose all 4 cards from that suit's 13 cards. Divide by the total ways to choose any 4 cards from 52.

Step-by-Step Solution

Ways to choose 4 cards from one suit: $\binom{13}{4}=715$ .

Number of suits: 4, so favorable outcomes $=4\times715=2860$ .

Total ways to choose any 4 cards from 52: $\binom{52}{4}=270725$ .

Probability $=\dfrac{2860}{270725}$ . …
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AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
$\dfrac{215}{221}$
›Reveal solutionSolution
The mean (expected value) of the number of kings among 2 drawn cards is found instantly by linearity of expectation: $E[X]=2\cdot\dfrac{4}{52}=\dfrac{2}{13}$ .

Concept and Intuition

Instead of building the full probability distribution of $X=$ number of kings (values 0, 1, 2) via hypergeometric probabilities and then computing $E[X]=\sum x\,P(X=x)$ , we can use linearity of expectation: write $X=I_1+I_2$ where $I_j$ is the indicator that the $j$ -th card is a king. Then $E[X]=E[I_1]+E[I_2]$ , and each indicator has expectation equal to the marginal probability a single card is a king, $\dfrac{4}{52}$ — this holds even though the two draws are dependent (without replacement), because expectation of a sum is always the sum of expectations.

Step-by-Step Solution

Let $I_1,I_2$ be indicators that the 1st, 2nd drawn card (in the pair) is a king.

By symmetry, $P(I_1=1)=P(I_2=1)=\dfrac{4}{52}=\dfrac{1}{13}$ (each individual card is equally likely to be any of the 52, of which 4 are kings).

$X=I_1+I_2$ , so $E[X]=E[I_1]+E[I_2]=\dfrac{1}{13}+\dfrac{1}{13}=\dfrac{2}{13}$ . …
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Q.A flashlight has $8$ batteries out of which $3$ are dead. If two batteries are selected without replacement and tested, the probability that both are dead is
(A) $\dfrac{33}{56}$
(B) $\dfrac{9}{64}$
(C) $\dfrac{1}{14}$
(D) $\dfrac{3}{28}$

Concept understanding — Hypergeometric Probability

Hypergeometric Probability: Drawing Without Replacement

The Intuition

The Precise Statement

Why This Formula Makes Sense

Worked Example

When to Use Hypergeometric …

Method: "Both are $X$ " without replacement — the multiplication theorem

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.A flashlight has 888 batteries out of which 333 are dead. If two batteries are selected without replacement and tested, the probability that both are dead is (A) 3356\dfrac{33}{56}5633​ (B) 964\dfrac{9}{64}649​ (C) 114\dfrac{1}{14}141​ (D) 328\dfrac{3}{28}283​

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Q.A flashlight has $8$ batteries out of which $3$ are dead. If two batteries are selected without replacement and tested, the probability that both are dead is
(A) $\dfrac{33}{56}$
(B) $\dfrac{9}{64}$
(C) $\dfrac{1}{14}$
(D) $\dfrac{3}{28}$

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