 Hello friends and how are you all doing today? The question says six coins are tossed simultaneously. Find the probability of getting three heads, no heads, at least one head. So here, in this question we are given n as six. Probability of getting a head is one by two right? So probability of getting a tail will be again one by two. Now that we will get by subtracting one by two from one. Now let X denotes the number of successes in the experiment. Then X varies from zero to six right? Now we know that we will find the probability that X is equal to r by using this formula that is n, c, r, p raised to the power r, q raised to the power n minus r where greater than equal to zero but less than equal to six. Therefore we need to find out probability of getting three heads, that is n, c, r, here n is six, c, r is three, p raised to the power r, q raised to the power n minus r where n is six and r is three. So on simplifying it we have six into five into four upon three into two into one, one upon two raised to the power three into one upon two raised to the power three. That gives us twenty by sixty four and on simplifying it we have it as five upon that is equal to five upon sixteen. So this is the answer to the first part. Proceeding with the second part. Here we need to find out the probability of getting no head. That means here our r is equal to zero right? So it will be. So on simplifying it we have one into one by two upon six giving us the answer as part. We need to find out the probability of one head minus probability of getting no head right? We found out above probability of getting no head as one by sixty four. So here the answer will be sixty three by sixty four right? So this is the answer to the last part solution. Hope you understood it. Have a nice day.