<p>if 5 five coins are flipped and all the different ways they could fall are listed, how many elements of this will contain more than 3 heads?
a.5 b.6 c.10 d. 16 e.32
answer is B</p>
<p>if there are known to be 4 broken transistors in a box of 12, and 3 are drawn at random, what is the probability that none of the 3 is broken
answer is .255</p>
<p>4 men, A,b, c, d line up in a row
what is the probability that man a is at either end of the row.
answer is .5</p>
<p>I tried doing binompdf and stuff for these since i took stats but i can't get the answers! wth.</p>
<ol>
<li><p>So the set needs to contain either 5 or 4 heads. Well, it will have 5 heads exactly once. But it can have four heads a number of times. Lets think of it differently. If it has four heads, that means there is exactly one tail, and so we need to count how many ways we can choose 1 tail out of the 5 possible slots, which is just 5 choose 1, or 5, so the total number of times is 1 + 5 = 6.</p></li>
<li><p>The probability the first one isn’t broken is 8/12. The probability the second isn’t broken WON’T BE 7/12, because we already took one out. So its probability is 7/11. The last probability is 6/10. Since these events are made linearly independent, the tota probability is 8/12 * 7/11 * 6/10 which is about .255.</p></li>
<li><p>A can be at either end, so two possible options, and there are a total of 4 places he can be, so the answer is just 2/4 = 1/2 = .5.</p></li>
</ol>
<p>Hope this helps!!</p>
<p>You don’t use binomial for 1. It’s strictly a combinations type problem.</p>
<p>i.e. In combinations: (5,4) + (5,5) = 5 + 1 = 6</p>
<p>For 3, it’s also a combinations problem.</p>
<p>You have four men with 2 slots at either end. So (4,2) = 12</p>
<p>Then you have 4! = 24 possible combinations of the four men.</p>
<p>So 12/24 = 0.5</p>
<p>My way is easier. :)</p>
<p>You can attack problems 1 and 3 in a number of ways . Only #2 is fairly set in stone.</p>