Anthony observes that when people are leaving class, about 50% turn left and about 50% turn right. anthony wants to estimate the probability that fewer than 3 of the first 4 students will turn left when leaving class tomorrow.
to do this, he lets h represent turning left when leaving class and t represent turning right. he then flips a coin 4 times, recording the results. he repeats this process for a total of 20 trials.

Hi! so:

H represent turning left when leaving class and T represent turning right,

Both of them have a 50% probability of happening.

and he wants to know the probability of less than 3 turn left so we wanna know the probability for only one to turn left and the probability for only two.

only one:

if you use the fact that we have a binomial distribution, then the Posibility that only one turns left is:

so P(1)= (4!/3!)*= 4*0.0625 = 0.25

only two:

with the same reasoning.

P(2) = 4!/(2!*2!)*0.0625=6*0.0625 = 0.375

here you can se that the probability is bigger because here are mas combinations that in the previous case.

So the total probability is 0.25 + 0.375 = 0.625.

The fact that he repeats the test n times doesnt change nothing, because he already know that you have a 50/50 chance of turning left of righth

5x+10 is the correct one

hope this

it’s supposed to be -4+45=41

step-by-step explanation: