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Assume the rancom variable X has binomial distribution with the given probability of obtaining _ success Find the following probability: given the number of trials...

Question

Assume the rancom variable X has binomial distribution with the given probability of obtaining _ success Find the following probability: given the number of trials and the probability of obtaining decimal places success Roundyour answer t0 foUrP = 15," = 8,p = 0.8

Assume the rancom variable X has binomial distribution with the given probability of obtaining _ success Find the following probability: given the number of trials and the probability of obtaining decimal places success Roundyour answer t0 foUr P = 15," = 8,p = 0.8



Answers

For the given probability of success $p$ on each trial, find the probability of $x$ successes in $n$ trials.
$$
x=4, n=8, p=0.7
$$

In this problem, we need to determine the probability of X successes in the an independent trials of a binomial probability experiment. Now we need to return on the probability of the number of successes being less than or equal to four. So this will be the probability of the number of successes being either zero or 1 Or two or 3 or four. Now all of these outcomes are disjointed. So using the addition law, probability this will be P0 must be one The speech to let's be three, let's be four. Now, using the formula for the probability of X successes and a binomial probability experiment, the value of P zero with the N C X at 12 0 times P, which is 0.35 to the power X. Which is zero times one minus P. So one minus 0.350 point 65 Keep our end minus X. So that Australian minus zero with just 12. Similarly, P one is 12 C 10.35 to the power 10.65 to the power 11, P two will be 12 C 20.35 to the power to time 0.65 to the power 10, P three is 12 C 30.35 to the power 30.65 to the Power nine. And P four will be 12 C 40.35 to the power four times 0.65 to the power eight. Now the value of this will be approximately 0.00 5 7 plus 0.0368 plus 0.1088 Plus 0.19 by four Plus 0.2367. And the value of this is equal to 0.5834. Hence the required probability is approximately equal to 0.5834.

Okay, this kind of probability you need to be able to use combinations and, um, communication. No, you don't need permutations, But let me assure you the form you like you could be using probability of something happening. It's going to be a combination because there's a much different ways it could happen. We just want four successes in a child's. We don't care when or when they happen near what order, so it's going to be whatever that number is. Times Pete the Eggs Times Q to the n minus X. This is all looks very scary, but we'll go ahead and break this down in just a second. Ex his how many times you have success. So in this case, exes who thinks he's going to be four and equals eight, or the number of total number of trials and hot Let's see P. The probability of success is your point. Dream and cue. The probability of not having a success is one minus 10.3 which 0.7 So we just need to figure out how do you plug all about into this equation we have here? Yes, we have eight C four and see times 0.3 to the fourth power times 0.7 to the fourth forward because this is a four. Because we have eight minus four, which just for this number here, if you forgot how to compute, um, combinations, permutations, stuff, combinations. This is going to be eight factorial all over eight, minus four factorial. Th I remember in two minutes for factorial tamed for a factorial. So this is going to be eight times. Seven times, six times, five times, four times, three times, two times one all over eight times seven times six four time, Three times, Two times, one times, four times, three times, two times one cancels. So we just need to type this into our calculators. Eight times. Seven times, six times five divided by four times, three times, two times one, which ends up being 70. So put that in to get 70 times 0.3 to the fourth Power times 0.7 to the fourth power to get the final answer Uh, 0.1361

In this problem we need to determine the probability of X successes in the n independent trials of a binomial probability experiment. Now the formula for the probability of X successes in an independent trials in a binomial probably the experiment is n c x times P to the power X times one minus P to depart and minus X. The value of N is given to be 15. The value of X is given to be tripped. The value of P is given to be 0.85 So we have 0.85 to the power 12 and we have one minus 0.85 to the power and minus X. So that is 15 minus 12. So we have 15 c. 12 which is equal to the green factorial divided by 12 factorial. That's 15 -12 which is three factory. But this we have 0.85 to the power 12 And 1 -0.85 which is 0.15 to the bar 15 -12, which is three. And the value of this is approximately equal to zero 21843. Hence the required probability is approximately equal to 0.21843.

No this problem we have been given that in his 12th p is 0.35 and we want it to be less than are able to form. We want to find that probability now for for binomial variable the probability of X. Is equal to end choose X. Mhm times P. E. To the X. That was one minus P to the n minus X. The probability X is less than 8.4 zero. The probability of zero plus the probability of one. What's the probability of two? Also the probability of three lost. The probability of four. Now finding each of these here, the probability of zero. It's 12 to 0 times 0.35 20 comes 0.65 to the 12th. I'm probably of one. It's 12 choose one times 0.35 to the one times 0.65 to the 11th. Yeah, probability of two is 12, choose to 5.35 square times 0.65 to the 10th. Probably a three 12 to 3 0.35 to the third point search five to the ninth. The probability of four it's 12 to 4 times 0.35 to the fourth times 0.65 to the eighth. Mhm. We have all these probabilities. It's just a matter of actually putting all these in and actually calculating all of this. You just need to calculate each of these individually and you should be able to just type each of these in your calculator when you find all the G. P 0.6096 That is our probability 0.6096


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