GO Classes DPP | GATE DA | Probability | Selection Without Replacement

Answer 1

There are $6$ cards labeled $1,2,3,4,5,6$.

Two cards are drawn one at a time, so order matters.

There are $6 \times 5 = 30$ ordered ways to draw two cards.

We want the first card to be exactly $2$ greater than the second card.

So if the second card is $1$, the first must be $3$.
If the second card is $2$, the first must be $4$.
If the second card is $3$, the first must be $5$.
If the second card is $4$, the first must be $6$.

The favorable ordered pairs are $(3,1), (4,2), (5,3), (6,4)$

So there are $4$ favorable outcomes.

Thus the probability is $\dfrac{4}{30} = \dfrac{2}{15}.$

Answer: $\boxed{\text{B. } \dfrac{2}{15}}$

Answer 2

Mistake: Favorable event space ordered pairs are flipped. That is wrong. Same result.
Answer: B

Answer 3

Given:

• A deck consists of $6$ cards carrying distinct numbers from $1$ to $6$.
• Two cards are removed one at a time.

To find:

The probability that the number on the first card is exactly $2$ greater than the number on the second card.

Step 1: Find the total number of possible outcomes in the sample space $n(S)$.

Since the cards are drawn one at a time without replacement, the order in which they are drawn matters.

• Number of choices for the first card = $6$
• Number of choices for the second card = $5$

Total possible outcomes $n(S) = 6 \times 5 = 30$

Step 2: Find the number of favorable outcomes $n(E)$.

Let the first card drawn be $x$ and the second card drawn be $y$.

The condition given is $x = y + 2$.

Since the cards are numbered $1$ to $6$, the possible valid pairs $(x, y)$ that satisfy this condition are:

• $(3, 1)$
• $(4, 2)$
• $(5, 3)$
• $(6, 4)$

The total number of favorable outcomes $n(E) = 4$.

Step 3: Calculate the probability.

\begin{align*}

P(\text{Event}) &= \frac{n(E)}{n(S)} \\

&= \frac{4}{30} \\

&= \frac{2}{15}

\end{align*}

Correct Option: B. $\frac{2}{15}$