Basic Probability
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Drexel University *
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310
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Mathematics
Date
Apr 3, 2024
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docx
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Uploaded by ndk35264519
Nency Kathiriya Basic Probability
1) You are attempting to determine whether a coin is fair with fair defined
having a 50% chance of landing heads up and a 50% chance of landing tails
up. a. You toss the coin four times and the result is 4 heads. What is the
probability of this result if the coin is fair? b. You tossed the coin three times
and the result is 3 heads. What is the probability of a head in the next toss of
the coin if the coin is fair? c. You tossed the coin seven times and the result is
7 heads. Can you conclude with certainty that the coin is not fair? Justify your
response. A). Each toss is independent, and the probability of getting heads each time
is 0.5. The probability of getting heads in a row is 0.5*4, which is 0.0625 or
6.25%. B). If the coin is fair, the probability of getting heads on the next toss is still
0.5. Each toss is independent, and past outcomes don't influence future
ones.
C). While getting 7 heads in 7 tosses may seem unlikely if the coin is fair
(probability is 0.5*7, which is about 0.0078 or 0.78%), it doesn't necessarily
mean the coin is not fair. Coin tosses are subject to randomness, and
occasionally getting a sequence of heads or tails is possible even with a fair
coin. To conclude with certainty that the coin is not fair, you would need a more
extensive and controlled experiment, such as many tosses with consistent
deviation from the expected 50% heads and 50% tails outcome.
2) Drexel University is conducting a randomized experiment to compare an
herbal remedy with a placebo for treating RSV infections. The health clinic
selects four student volunteers, two of whom are male (Shi and Mateo) and
two are female (Anika and Malaika). Two of the volunteers will be chosen at
random to receive the herbal remedy while two will be chosen at random to
receive placebo. a. What is the sample space for those receiving the herbal
remedy? b. What is the probability of the event that the sample chosen to
receive the herbal remedy will consist of 1 male and one female?
- a) The sample space for those receiving the herbal remedy is
S = {(B1,B2), (B1,G1), (B1,G2), (B2,G1), (B2,G2), (G1,G2) }
b) Probability that the event that the sample chosen to receive the herbal
remedy will consist of 1 male and one female is,
P(A) =m/n = 4/6 =2/3 =0.66
3) A jury of 12 people is chosen for a trial. The defense attorney claims that
the jury must have been chosen in a biased manner, because 50% of the
city’s adult residents are female, yet the jury contains no women. a.
Assuming that the jury was randomly chosen from the population, what is
the probability that the jury would have no women members? b. Assuming
the jury was randomly chosen from the population, what is the probability
that the jury would have at least one woman member?
- a) The probability that the jury would have no women members.
P(X=0) =(12 0)(0.5)^12=(0.5)^12=0.000244
- The probability that the jury would have no women members is 0.000244.
b) The probability that the jury would have at least woman member.
P(X≥1) =1−P(X<1)=1−P(X=0)=1−(12 0)(0.5)^12,=1−0.000244=0.999756
- The probability that the jury would have at least woman member is
0.999756.
4) There is a box containing 8 white balls, 5 black balls, and 9 red balls. If 2
balls are drawn one at a time from the box and neither is replaced, find the
probability that: a. both balls will be white. b. the 1st ball will be white and
the 2nd ball black. c. if a 3rd ball is drawn, find the probability that the 3
balls will be drawn in the order: white then black then red.
a. Probability that both balls will be white:
P(both white)=Number of ways to choose 2 white balls /Total number of ways to choose 2 balls
P(both white)={8ch∞se2}/{22ch∞se2}
P(both white)=(8!/2!6!)/(22!2!20!_
P(both white)=28/231
b. Probability that the 1st ball will be white and the 2nd ball black:
P(1st white, 2nd black)=Number of ways to choose 1 white and 1 black ball/total number of ways to choose 2 balls
P(1st white, 2nd black)={8ch∞se1}
⋅
{5ch∞se1}/{22ch∞se2}
P(1st white, 2nd black)=(8!/1!7!
⋅
5!/1!4!)/(22!2!20!)
P(1st white, 2nd black)=80/231
c. Probability of drawing 3 balls in the order white, black, red:
P(white, black, red)=Number of ways to choose 1 white, 1 black, and 1 red ball in that order/ Total number of ways to choose 3 balls
P(white, black, red)={8ch∞se1}
⋅
{5ch∞se1}
⋅
{9ch∞se1}/{22ch∞se3}
P(white, black, red)=(8!/1!7!)
⋅(
5!/1!4!)
⋅(
9!/1!8!)/(22!/3!19!)
P(white,black,red) = 180/1771
5) Suppose a die has been loaded so that the one face lands uppermost 4
times as often as any other face while all the other faces occur equally often.
What is the probability of the die landing with the 2 face up on a single toss?
What is the probability of the die landing with the 1 face up on a single toss?
- Given that one face lands uppermost 4 times as often as any other face while all the other faces occur equally often, let's denote the probability of the loaded face as P(loaded) and the probability of any other face as P(other).
Let P(other)=x. Then P(loaded)=4x.
The sum of all probabilities must be equal to 1:
P(loaded)+5
⋅
P(other)=1
Substitute the values:
4x+5x=1
9x=1
x=1/9
Now, we can find the probabilities of each face:
1. Probability of landing with the 2 face up P(2):
P(2)=P(other)=1/9
2. Probability of landing with the 1 face up P(1):
P(1)=P(loaded)=4
⋅
1/9=4/9
So, the probability of the die landing with the 2 face up is 1/9, and the probability of the die landing with the 1 face up is 4/9.
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