Probability of Dependent Events | MathHelp.com - By MathHelp.com
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| 00:00 | Andrew has eight black socks and four brown socks in | |
| 00:04 | his sock drawer and it's too dark to tell the | |
| 00:06 | colours apart . If he chooses one sock and puts | |
| 00:10 | it on , then chooses another . Sock . Find | |
| 00:13 | the probability of the following event . Pf brown than | |
| 00:16 | brown . In other words , were asked to find | |
| 00:20 | the probability that Andrew chooses a brown sock , then | |
| 00:23 | another brown sock . It's important to understand that when | |
| 00:27 | Andrew chooses the first suck , he puts it on | |
| 00:30 | , so when he chooses the second sock , there's | |
| 00:33 | one fewer sock in the drawer . Therefore the outcome | |
| 00:37 | of the first event affects the outcome of the second | |
| 00:40 | event , which means that these are dependent events . | |
| 00:43 | To find the probability of dependent events . We first | |
| 00:47 | find the probability of each event , then we multiply | |
| 00:50 | the probabilities together just like we did when finding the | |
| 00:53 | probability of independent events . However , we have to | |
| 00:57 | be very careful when finding the probabilities . So let's | |
| 01:01 | first find the probability that Andrew chooses a brown sock | |
| 01:05 | . What ? Since there are four brown socks in | |
| 01:14 | the drawer and there are eight plus four or 12 | |
| 01:18 | total socks in the drawer . The probability that Andrew | |
| 01:21 | chooses a brown sock is four 12 . Which reduces | |
| 01:27 | to 1/3 . Remember to always reduce if possible . | |
| 01:32 | When finding probability . Next let's find the probability that | |
| 01:36 | Andrew chooses a second . Brown sock . Yeah . | |
| 01:43 | Yeah . Well yes and here's where it gets tricky | |
| 01:49 | , remember that when Andrew chooses the first sock he | |
| 01:52 | puts it on . So if he has already successfully | |
| 01:55 | chosen a brown sock , there are no longer four | |
| 01:58 | brown socks in the drawer . There are three and | |
| 02:02 | there are now eight plus three Or 11 total socks | |
| 02:08 | in the drawer . So the probability that Andrew chooses | |
| 02:11 | a second brown sock is three 11th . Now to | |
| 02:18 | find the probability that he chooses a brown sock puts | |
| 02:21 | it on , then chooses another Brown sock . We | |
| 02:24 | multiply the probability that Andrew chooses a brown sock one | |
| 02:28 | third . Mhm . Times the probability that he chooses | |
| 02:35 | a second . Brown Sock 3 11th notice that we | |
| 02:43 | can cross cancel the three and 321 and one now | |
| 02:48 | multiplying across the numerator one times one is one , | |
| 02:52 | and multiplying across the denominators , one times 11 is | |
| 02:57 | 11 , so one third times 3/11 is 1/11 which | |
| 03:02 | means the probability that Andrew chooses a brown sock , | |
| 03:06 | puts it on , then chooses another . Brown sock | |
| 03:09 | is 1/11 . |
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