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A certain unfair coin lands on heads 1$/ 4$ of the time. This coin is tossed 6 times. Find the probabilities of getting the following.
At least 3 heads

Okay, This question asks us the probability of getting no more than three heads from this unfair coin in six toss. It's so we want the probability of no more than three. And this could happen by getting zero heads. One head, two heads, 43 heads. So we need to find all of these probabilities and add them together. So probability of zero heads Well, that's just the chance of not getting ahead happening six times in a row. And then for the 1st 1 we have six coins, and we want one of them to be ahead. And that had happens 1/4 of the time. Which means the other five flips have to be tales, and we could go through and write all of these probabilities out fun. Since we have access to calculators, this could be conveniently written in this form. The probability of getting less than or equal to a certain number is just the binomial CDF function on a calculator of six trials with a 1/4 success rate, and we want no more than three. So this saves this time and it's approximately 0.96 to 4, and that's our answer

In this problem that Q. is equal to one -P. Now since head appears first time and then even through then head can occur. I did it the second through fourth through 6th 2 and so on. Therefore we have To buy five equals 2. To get in the second room will be queuing to pee, has to get under food through Q. Q. At a still dale dale and then ahead you cube plus entropy next Or getting in the 6th. True for the first traffic will be deal and then I had and this goes up. So this is a geometric progression Post on rescue. P one the common ratio is you square. He says to buy five equals 2, curious one minus B In two p. This is 1 -4 1- B sq. So we get to buy five equals two. This comes us one may must be divided by two minus B. Four minus Group B equals two five minus. So I speak From here we get three pairs, 1 R. B. S., one or 2. The correct option is it. That's all.

Okay, This question asks us to find the probability of getting all heads in six tosses from a coin that has a 1/4 chance of coming up heads. So there's actually only one way to do this. And that's if we had that 1/4 chance of getting heads happening six times back to back. So this is one divided by four to the sixth. Or doing the math on this that's just one over 40 96 connects her answer.

Okay, This question wants us to find the probability of getting exactly three heads from this unfair coin out of six tosses. So we're going to use this binomial distribution where we have que successes out of end trials. So it's and choose k times the success probability to the k times, the failure probability to the n minus cake. So we want to know the probability of getting exactly three heads, which is out of the six tosses. We need three to be heads and the heads happened 1/4 of the time and we need that's happened three times. Which means that the 3/4 chance of coming up tails must happen the remaining three times. So plugging this in this is equal to 5 40 divided by 40 96. If you want to write, this is a fraction