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Number one, suppose a random process has a total of three basic outcomes, 0 sub 1, 072, and 073.
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In other words, its sample space is 0 .1, 0 sub 2, comma, 07 3.
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The probabilities of 07 2 and 07 3 are 0 .5 and 0 .4 respectively.
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Let e be the event consisting of the basic outcomes, 02 and 03.
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That is, e is said to occur if either 02 or 03 occurs.
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So the complement to e would be.
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So let's start with the probability of e.
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The probability of e is going to be 0 .5 or 0 .4.
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So that would be 0 .9.
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So the complement would be 1 minus 0 .9, which would be 0 .1.
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And that matches up with a.
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Number 2, 2 marbles are drawn at random and without replacement from a box containing two blue marbles and three red marbles.
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Determine the probability of observing the following events.
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So the probability of drawing two blue marbles, we start off with two out of the five that are blue.
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Then we have one out of four that are blue for a total of two out of 20, which would be 0 .1.
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For the probability of red blue, or we could say blue -red, so one of each, we'll have two times the probability of three.
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And i'm going to write this as a combined, so three times two in the numerator, five times four in the denominator.
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So we have two times 6 .20th, which would be 1220th, which is 0 .6.
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And the probability of red red would be 3 fifths times, excuse me, 2 .4s.
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So it would be 620th, and that would be 0 .3.
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Notice that those three probabilities add to be 1.
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That's what we should be looking for.
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Number 3, a club has 100 members, 40 of whom are lawyers, and 30 of whom are liars.
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A total of 60 members are neither lawyers nor liars.
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Out of the club's lawyers, what proportion are liars? oof, that's almost a tongue twister.
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So they recommend that we do a venn diagram, and i wholeheartedly agree.
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So we have the lawyer, we have the liar.
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We know there's 100 total, and we know that 40 wouldn't fit into either category.
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I'm sorry, not 40, 60, which was, i guess, nice.
02:33
60 would not fit in either category...