Probability

Let's say you have 6 cards numbered 1 to 6 in a box:

In your first draw, you pickup a card randomly and remove it from the box.

In second draw again you pickup a card randomly and remove it from the box but you put back the first card that you had drawn in the previous step.

Calculate the probability that during your third draw you will draw the same card again that you had drawn during your first draw.

#Combinatorics

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

20 percent.

Jeff Giff - 4 months, 1 week ago

Correct. While this is an easy problem, while learning probability students often think about this in terms of conditional probability.

Vikram Pandya - 4 months, 1 week ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$