 Hello and welcome to the session. Let's work out the following problem. It says a pair of dice is thrown seven times If getting the total of seven is considered as success, find the probability of no success at least six successes So let's now move on to the solution And let's first write the sample space for the experiment of throwing a pair of dice Since we are throwing a pair of dice There would be 36 possible outcomes These are one six first we add one one one two one three one four one five one six Similarly, we can write out all the outcomes So we have written all the possible outcomes and these are 36 and numbers now Let e be the event of getting a total of seven now this is considered as success the probability of Getting a total of seven would be six by 36 as there are six outcomes which are possible to Which are favorable to e that is which gives us a total of seven as one six two seven three four four three Five two and six one so these are six in numbers and the total number of outcomes are 36 So the probability of getting a total of seven is six by 36 that is one by six so the probability of not getting a total of seven will be one minus probability of getting a total of seven That is one minus one by Six that is five by six Now we are throwing aware of dice seven times so let X denotes the number of Successes in seven throws now X is a binomial variant since the number of throws are finite therefore X is a binomial variate right where N is seven and P probability of success is One by six now we know that the probability at X is equal to R for the binomial distribution is given by NCR P to the power R Q to the power N minus R where R goes From zero one to a pill and so you need to remember the binomial Distribution for this now here we have to find Probability of no success that is probability of X zero since X denotes the number of successes and seven throws and we have to find the probability of no success That means X takes the value as zero Now we'll apply This distribution here. So probability of X at zero will be seven C zero P to the power R that is one by six to the power zero into Q Q is five by six five by six to the power Seven minus zero So this is seven C zero One by six to the power zero is one Into five by six to the power seven and seven C zero is one. So this is Five by six to the power seven Now in the second part we have to find the probability of at least six successes probability of At least six successes that means There could be six successes or more than six successes in seven throws. That means X takes the values Greater than or equal to six and since there are seven throws. So it would be probability of X Equal to six plus probability of X equal to seven Now again probability of X is equal to six will be seven C six one by six to the power six five by six to the power seven minus six plus seven C seven One by six to the power seven five by six to the power seven minus seven now Seven C seven and this is one by six to the power six and five by six to the power one Plus seven C seven is one into one by six to the power seven and five by six to the power zero So this is equal to 7 into 1 by 6 to the power 6 5 by 6 to the power 1 plus 1 by 6 to the power 7. So now here we can take 1 by 6 to the power 7 common. So we take 1 by 6 to the power 7 as common. So we have inside. Now since we are taking 1 by 6 to the power 7 common here we have 7 into 5 plus 1. So this is equal to 1 by 6 to the power 7 into 35 plus 1. So this is 1 by 6 to the power 7 into 36. Again this is 1 by 6 to the power 7, 36 is 6 to the power 2. So this is equal to 1 by 6 to the power 5. Since here 1 by 6 to the power 2 gets cancelled with 1 by 6 to the power 2. This gets cancelled with this and we have 1 by 6 to the power 5. So the probability of no success is 5 by 6 to the power 7 and probability of at least 6 successes is 1 by 6 to the power 5. So this completes the question and the session. Bye for now. Take care. Have a good day.