Econ 41 / Stats Homework Question.

Colleges and Universities A-Z University of California - Los Angeles
1

<p>For any of you in Econ 41 or are good in stats, I have a problem for you that I need help on...</p>

<p>Two playing cards are drawn successively and without replacement from a standard deck. Compute the probability of drawing...
A heart on the first draw, an ace on the second draw. (note that a heart can be drawn by getting the ace of hearts or one of the other 12 hearts)</p>

<p>I was thinking of setting up two probabilities and adding them, something like this...
P(Ace of Hearts on first draw) x P(Ace on second draw)
+
P(Heart thats not the Ace of Hearts on the first draw) x P(Ace on second draw)</p>

<p>Which would then give me...</p>

<p>(1/52)(3/51) + (12/52)(4/51) = .0192</p>

<p>Does this seem correct?</p>

<p>Thanks</p>

2

<p>That’s what I would do, but then again I’m retaking it so don’t take it from me.</p>

3

<p>Here’s a more rigorous way to look at it (‘|’ means “given”):</p>

<p>P(1st is heart AND 2nd is ace) = P(first is heart) * P(second is ace|first is heart) = P(first is heart) * [P(first is ace) * P(second is ace|first is heart and first is ace) + P(first is not ace) * P(second is ace|first is heart and first is not ace)] = (1/4) * [(1/13) * (1/17) + (12/13) * (4/51)] ~= 0.0192</p>

<p>So, your approach works.</p>

4

<p>can anyone post the solutions of HW, please?</p>