 Hello and welcome to this session. In this session we are going to discuss formation of sample spaces using list, table and tree diagrams. When an experiment involves more than one event, then the sample space can be described using list, table and tree diagram. Let us suppose we toss a coin and spin a spinning wheel together. Here coin has two outcomes head and tail and spinner has eight outcomes bubbles 1, 2, 3, 4, 5, 6, 7 and 8. Now we will describe this event of tossing a coin and spinning a wheel together with the help of a list. Now we list all possible outcomes that is if head comes in tossing a coin any number can come on spinner. So we have outcomes h1, h2, h3, h4, h5, h6, h7 and h8. Similarly if tail comes in tossing a coin then again spinner can have any number possible outcomes h1, h2, h3, h4, h5, h6, h7 and h8. So sample space would be equal to the set containing elements h1, h2, h3, h4, h5, h6, h7, h8, t1, t2, t3, t4, t6, t7 and t8. This is the required list. Next we have table. Let us form a table for the outcomes. The rows represent the outcomes for the coin that is head and tail and the columns represent the outcomes for the spinner that is 1, 2, 3, 4, 5, 6, 7 and 8. Now we have written the combination in the table cells corresponding to respective row and column elements. See the shared cell in the table h5. Here the corresponding row outcome, column outcome is 5. So our element for cell is h5 which shows the appearance of 5 on the spinner that appears on the coin will be the required outcome. Now we are going to discuss tree diagram, the tree diagram and item that branches into two or more each of which branch into two or more and so on. It looks like a tree with trunk multiple branches. Here the first set of branches is the outcome of tossing a coin that is a head or a tail. For each of these outcomes there is a second set of branches representing the outcomes of a spinner that is 1, 2, 3, 4, 5, 6, 7 and 8. Each of these representations show there are 16 outcomes with the help of any of the above representations we can identify all the possible outcomes of the event. Let us take an example. If we want to identify all the outcomes which represent the event, a head and an even number. So from the sample space we can find the favorable outcomes for the given event. And a sample space is given by the set containing elements h1, h2, h3, h4, h5, h6, h7, h8, t1, t2, t3, t4, t5, t6, t7 and t8. Let event a be equal to getting a head and even number. We see that we have four combinations of head and even number. So event a is equal to the set containing elements h2, h4, h6, h8. This completes our session. Hope you enjoyed this session.