グローバルビジネスで役立つ数学（92）エースを引く確率（英語版）

Hi there!

Today, let’s quickly get down to a word problem related to odds.

From a deck of 52 cards we draw two cards one by one without replacement. Find the probability that both cards are Aces. Don’t forget to reduce the denominator and numerator of a fraction by a common divisor. (Source: A foreign mathematical web site)

Here is my answer to the problem.

It’s loud and clear that \(n(S) = 52\).

Let \(A\) = event of drawing first Ace.
\(n(A) = 4\,\Rightarrow P(A) = \frac{4}{52}\)

Let B = event of drawing second Ace when first is drawn.
\(n(B) = 3\).

Also we have drawn a card already so, \(P(B) = \frac{3}{51}\).

So the probability of occurrence of both \(A\) & \(B\) is given by the product of individual probabilities.

\(P(A\cap B) = P(A).P(B)\)
\(= \frac{4}{52}・\frac{3}{51}\)
\(= \frac{1}{13}・\frac{1}{17}\)
\(= \frac{1}{221}\)

If anyone found the source of the above question, I would be more than happy if you could let me know the exact source of the problem. I will lose no time in describing the source without any delay. Thanks.

Stay tuned, and expect to see my next post.

Keep well.

Frank Yoshida

￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣
【グローバルビジネスで役立つ数学】でもっと学習する b＾＾)
【参考図書】『もう一度高校数学』（著者：高橋一雄氏）株式会社日本実業出版社
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