Probability Help

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1

<p>I was looking at the SAT question of the day and trying to solve it. "Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?"</p>

<p>The first thing I thought to do was to find the probability of each individual event occurring and then combine and multiply.</p>

<p>Probability of Male 3/5
Probability of Office 3/5</p>

<p>Probability of 2nd Male 2/4
Probability of Office 2/4</p>

<p>Probability of Female 2/3
Probability of Office 1/3</p>

<p>(3/5 * 3/5) + (2/4 * 2/4) + (2/3 * 1/3)
9/25 + 4/16 + 2/9 = 0.8322222....</p>

<p>The correct answer however is 3/5. I realize this can be achieved by not subtracting one from the denominator each time and getting:</p>

<p>(3/5 * 3/5) + (2/5 * 2/5) + (2/5 * 1/5) = 3/5</p>

<p>But why wouldn't you subtract one from the denominator after each time? That person or office space is no longer available after it has been chosen or filled, right?</p>

2

<p>There are 5C3 = 10 ways to pick a subset of three employees to assign to offices. How many ways can we pick two men, one woman? We can pick two men in 3C2 = 3 ways, and pick one woman in 2 ways. Therefore there are 3*2 = 6 ways to choose two men and one woman, 6/10 = 3/5.</p>

<p>As for your solution, I’m not sure why you’re multiplying 3/5 twice. We pick a male at random and give him a workspace at random, but what if the person picked is not male?</p>

<p>An equally valid solution is to consider the 120 permutations of the 5 employees and assign cubicles to the last two employees. We want the last two employees to contain exactly one man, one woman. The number of ways to do this is 3<em>2</em>2*6 = 72 (pick one man, one woman, 2 ways to order them, 6 ways to order the remaining 3 employees), and 72/120 = 3/5.</p>