Probability of Independent Events | MathHelp.com - By MathHelp.com
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| 00:0-1 | in this example we're given that a coin is tossed | |
| 00:03 | twice and were asked to find the probability of getting | |
| 00:06 | heads and heads . It's important to understand that the | |
| 00:10 | two coin tosses are called independent events because the outcome | |
| 00:15 | of the first coin toss does not affect the outcome | |
| 00:18 | of the second coin toss . To find the probability | |
| 00:22 | of independent events . We first find the probability of | |
| 00:25 | each event . Then we multiply the probabilities together . | |
| 00:30 | So in this problem P of heads and heads means | |
| 00:34 | the same thing as p of heads . Okay , | |
| 00:40 | Times p of heads . Yeah . So let's start | |
| 00:48 | by finding the probability of getting heads on our first | |
| 00:51 | coin toss . Since there's only one way to get | |
| 00:54 | heads in a coin toss and there are two possible | |
| 00:57 | outcomes , heads or tails . The probability of getting | |
| 01:01 | heads on our first coin toss is one half next | |
| 01:06 | let's find the probability of getting heads on our second | |
| 01:09 | coin toss which would also be one half . So | |
| 01:13 | we have one half times one half Now multiplying across | |
| 01:19 | the numerator one times 1 is one and multiplying across | |
| 01:25 | the denominators two times two is four , So one | |
| 01:30 | half times one half is 1/4 , which means that | |
| 01:34 | the probability of tossing a coin twice and getting heads | |
| 01:37 | and heads is 1/4 . |
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