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: Hypergeometric Probability — drawing without replacement from a finite set with two types.

Step 1: Total ways to choose any 2 batteries from 8:

$\binom{8}{2} = 28$.

Step 2: Ways to choose 2 dead batteries from the 3 dead ones:

$\binom{3}{2} = 3$.

Step 3: Probability both are dead: …

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

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

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

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

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

Use this when you take two (or more) items one after another without replacement and want them all to be of a particular type. The draws are dependent, so use the conditional multiplication theorem.

Steps

Step 1: Note the type-count and total before the first draw.

Let there be $d$ items of the wanted type out of $N$ total.

Step 2: Probability the first draw is of that type.

Step 3: Update the counts, then find the conditional probability of the second.

After removing one wanted item, $d-1$ remain out of $N-1$:

$P(\text{second} \mid \text{first}) = \frac{d-1}{N-1}$ …

Common Mistakes

Mistake 1: Treating the draws as independent (with replacement), $\tfrac38\times\tfrac38 = \tfrac{9}{64}$.

Why it's wrong: the batteries are selected without replacement, so the second draw sees one fewer dead battery and one fewer total. This gives the distractor $\tfrac{9}{64}$. Correct approach: reduce both counts — $\tfrac38\times\tfrac27 = \tfrac{3}{28}$.

Mistake 2: Reducing the dead count but not the total (or vice versa). …