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Sagot :
Alright, let's go through the geometric probability distribution problem step-by-step.
### (a) Probability [tex]\( P(3) \)[/tex]
Given:
- [tex]\( p = 0.524 \)[/tex] (Shaquille O'Neal's success rate)
- [tex]\( x = 3 \)[/tex] (we want the probability he makes his third attempt after missing two)
The probability [tex]\( P(x) \)[/tex] for a geometric distribution is given by:
[tex]\[ P(x) = p (1 - p)^{x-1} \][/tex]
So, for [tex]\( x = 3 \)[/tex]:
[tex]\[ P(3) = 0.524 \times (1 - 0.524)^{3-1} \][/tex]
From the result obtained:
[tex]\[ P(3) = 0.118725824 \][/tex]
### (b) Probability Distribution for [tex]\( x = 1, 2, \ldots, 10 \)[/tex]
We need to construct a probability distribution for [tex]\( x = 1 \)[/tex] to [tex]\( x = 10 \)[/tex]. The probability distribution [tex]\( P(x) \)[/tex] is:
1. For [tex]\( x = 1 \)[/tex]:
[tex]\[ P(1) = 0.524 \][/tex]
2. For [tex]\( x = 2 \)[/tex]:
[tex]\[ P(2) = 0.524 \times (1 - 0.524)^1 = 0.249424 \][/tex]
3. For [tex]\( x = 3 \)[/tex]:
[tex]\[ P(3) = 0.524 \times (1 - 0.524)^2 = 0.118725824 \][/tex]
4. For [tex]\( x = 4 \)[/tex]:
[tex]\[ P(4) = 0.524 \times (1 - 0.524)^3 = 0.056513492224 \][/tex]
5. For [tex]\( x = 5 \)[/tex]:
[tex]\[ P(5) = 0.524 \times (1 - 0.524)^4 = 0.026900422298624 \][/tex]
6. For [tex]\( x = 6 \)[/tex]:
[tex]\[ P(6) = 0.524 \times (1 - 0.524)^5 = 0.012804601014145 \][/tex]
7. For [tex]\( x = 7 \)[/tex]:
[tex]\[ P(7) = 0.524 \times (1 - 0.524)^6 = 0.006094990082733 \][/tex]
8. For [tex]\( x = 8 \)[/tex]:
[tex]\[ P(8) = 0.524 \times (1 - 0.524)^7 = 0.002901215279381 \][/tex]
9. For [tex]\( x = 9 \)[/tex]:
[tex]\[ P(9) = 0.524 \times (1 - 0.524)^8 = 0.001380978472985 \][/tex]
10. For [tex]\( x = 10 \)[/tex]:
[tex]\[ P(10) = 0.524 \times (1 - 0.524)^9 = 0.000657345753141 \][/tex]
Thus, the probability distribution is:
[tex]\[ [0.524, 0.249424, 0.118725824, 0.056513492224, 0.026900422298624, 0.012804601014145, 0.006094990082733, 0.002901215279381, 0.001380978472985, 0.000657345753141] \][/tex]
### (c) Mean of the Distribution
The mean [tex]\( \mu \)[/tex] of a geometric distribution is given by:
[tex]\[ \mu = \frac{1}{p} \][/tex]
Using [tex]\( p = 0.524 \)[/tex]:
[tex]\[ \mu = \frac{1}{0.524} \][/tex]
From the result obtained:
[tex]\[ \mu = 1.9083969465648853 \][/tex]
### (d) Comparison of the Mean with [tex]\( \frac{1}{p} \)[/tex]
We have already computed the mean as [tex]\( \frac{1}{p} = 1.9083969465648853 \)[/tex].
So, the mean calculated in part (c) [tex]\( \mu = 1.9083969465648853 \)[/tex] is indeed equal to [tex]\( \frac{1}{p} \)[/tex]. This confirms that the mean of a geometric distribution is [tex]\( \frac{1}{p} \)[/tex].
In conclusion, the expected number of free throw attempts Shaquille O'Neal would make before observing a successful free throw is approximately 1.908.
### (a) Probability [tex]\( P(3) \)[/tex]
Given:
- [tex]\( p = 0.524 \)[/tex] (Shaquille O'Neal's success rate)
- [tex]\( x = 3 \)[/tex] (we want the probability he makes his third attempt after missing two)
The probability [tex]\( P(x) \)[/tex] for a geometric distribution is given by:
[tex]\[ P(x) = p (1 - p)^{x-1} \][/tex]
So, for [tex]\( x = 3 \)[/tex]:
[tex]\[ P(3) = 0.524 \times (1 - 0.524)^{3-1} \][/tex]
From the result obtained:
[tex]\[ P(3) = 0.118725824 \][/tex]
### (b) Probability Distribution for [tex]\( x = 1, 2, \ldots, 10 \)[/tex]
We need to construct a probability distribution for [tex]\( x = 1 \)[/tex] to [tex]\( x = 10 \)[/tex]. The probability distribution [tex]\( P(x) \)[/tex] is:
1. For [tex]\( x = 1 \)[/tex]:
[tex]\[ P(1) = 0.524 \][/tex]
2. For [tex]\( x = 2 \)[/tex]:
[tex]\[ P(2) = 0.524 \times (1 - 0.524)^1 = 0.249424 \][/tex]
3. For [tex]\( x = 3 \)[/tex]:
[tex]\[ P(3) = 0.524 \times (1 - 0.524)^2 = 0.118725824 \][/tex]
4. For [tex]\( x = 4 \)[/tex]:
[tex]\[ P(4) = 0.524 \times (1 - 0.524)^3 = 0.056513492224 \][/tex]
5. For [tex]\( x = 5 \)[/tex]:
[tex]\[ P(5) = 0.524 \times (1 - 0.524)^4 = 0.026900422298624 \][/tex]
6. For [tex]\( x = 6 \)[/tex]:
[tex]\[ P(6) = 0.524 \times (1 - 0.524)^5 = 0.012804601014145 \][/tex]
7. For [tex]\( x = 7 \)[/tex]:
[tex]\[ P(7) = 0.524 \times (1 - 0.524)^6 = 0.006094990082733 \][/tex]
8. For [tex]\( x = 8 \)[/tex]:
[tex]\[ P(8) = 0.524 \times (1 - 0.524)^7 = 0.002901215279381 \][/tex]
9. For [tex]\( x = 9 \)[/tex]:
[tex]\[ P(9) = 0.524 \times (1 - 0.524)^8 = 0.001380978472985 \][/tex]
10. For [tex]\( x = 10 \)[/tex]:
[tex]\[ P(10) = 0.524 \times (1 - 0.524)^9 = 0.000657345753141 \][/tex]
Thus, the probability distribution is:
[tex]\[ [0.524, 0.249424, 0.118725824, 0.056513492224, 0.026900422298624, 0.012804601014145, 0.006094990082733, 0.002901215279381, 0.001380978472985, 0.000657345753141] \][/tex]
### (c) Mean of the Distribution
The mean [tex]\( \mu \)[/tex] of a geometric distribution is given by:
[tex]\[ \mu = \frac{1}{p} \][/tex]
Using [tex]\( p = 0.524 \)[/tex]:
[tex]\[ \mu = \frac{1}{0.524} \][/tex]
From the result obtained:
[tex]\[ \mu = 1.9083969465648853 \][/tex]
### (d) Comparison of the Mean with [tex]\( \frac{1}{p} \)[/tex]
We have already computed the mean as [tex]\( \frac{1}{p} = 1.9083969465648853 \)[/tex].
So, the mean calculated in part (c) [tex]\( \mu = 1.9083969465648853 \)[/tex] is indeed equal to [tex]\( \frac{1}{p} \)[/tex]. This confirms that the mean of a geometric distribution is [tex]\( \frac{1}{p} \)[/tex].
In conclusion, the expected number of free throw attempts Shaquille O'Neal would make before observing a successful free throw is approximately 1.908.
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