 Hello and welcome to the session. I am Shashi and I am going to help you with the following question. Question says, find the probability of throwing at most two sixes in six throws of a single die. Let us now start with the solution. Now we know that repeated throws of a single die are Bernoulli trials. So we can write the repeated throws of a die are Bernoulli trials. Now let us recall that Bernoulli trials are finite in number. They are independent. They have two outcomes success or failure and the probability of success remains same in each trial. Now throwing a die six times represents Bernoulli trials. Now let us assume that X denotes the number of sixes in an experiment of six trials. So we can write let X denotes the number of sixes in an experiment of six trials. Now we know probability of getting six in a single throw of die is equal to one upon six and probability of not getting six in a single throw of die is equal to one minus one upon six which is further equal to five upon six. Now let us recall that for binomial distribution P and P probability of X successes is equal to Ncx q raised to the power N minus X, P raised to the power X where X is equal to 0, 1, 2, till N. Now we have to find the probability of throwing at most two sixes. Now we can write probability of at most two sixes is equal to probability of X less than equal to 2. Now probability of getting successes less than equal to 2 is equal to probability of zero success plus probability of one success plus probability of two successes. Now we can find the probability of zero success by using this formula. We know total number of trials made is equal to six so value of N is equal to six. So here we can write six is zero value of q is five upon six. So here we can write five upon six raised to the power N minus X value of N is six and value of X is zero. So six minus zero is equal to six only. Now value of P is one upon six. So here we will write one upon six for P and value of X is zero here. So here we will write zero. Similarly we can find probability of getting a single success. Here we will substitute one for X and we get six C one multiplied by five upon six raised to the power five multiplied by one upon six raised to the power one. Now we will find probability for two successes. It is equal to six C two five upon six raised to the power four multiplied by one upon six raised to the power two. Remember that we have written these plus signs as it is. Now solving this step further we get we know six is zero is equal to one and anything raised to the power zero is equal to one. So here we get five upon six raised to the power six. Now simplifying this term we get six multiplied by five upon six raised to the power five multiplied by one upon six. Now here this six and this six will get cancelled. Now simplifying this term we get fifteen multiplied by five upon six raised to the power four multiplied by one upon thirty six. Now taking five upon six raised to the power four common in these three terms we get five upon six raised to the power four multiplied by twenty five upon thirty six. Plus five upon six plus fifteen upon thirty six. Now adding these three terms we get five upon six raised to the power four multiplied by seventy upon thirty six. Now we will cancel common factor two from numerator and denominator both here. And we get five upon six raised to the power four multiplied by thirty five upon eighteen. So our required answer is probability of throwing that most two sixes in six throws of a single die is equal to thirty five upon eighteen multiplied by five upon six raised to the power four. This completes the session. Hope you understood the solution. Take care and have a nice day.