Respuesta :
Answer:
Required probability = 0.3571
Step-by-step explanation:
We are given that a homeowner plants 6 bulbs selected at random from a box containing 5 tulip bulbs and 4 daffodil bulbs.
Number of ways of selecting 2 daffodil bulbs from total of 4 daffodil bulbs is given by = [tex]^{4} C_2[/tex] = [tex]\frac{4!}{2!*2!}[/tex] = 6
Number of ways of selecting 4 tulip bulbs from total of 5 tulip bulbs is given by = [tex]^{5} C_4[/tex] = [tex]\frac{5!}{4!*1!}[/tex] = 5
And Number of ways of selecting 6 bulbs from total of 9 bulbs in the box is given by = [tex]^{9} C_6[/tex] = [tex]\frac{9!}{6!*3!}[/tex] = 84
So, probability that he planted 2 daffodil bulbs and 4 tulip bulbs = [tex]\frac{^{4} C_2*^{5} C_4}{^{9} C_6}[/tex]
= [tex]\frac{6*5}{84}[/tex] = 30/84 = 0.3571
Therefore, required probability = 0.3571.
Using the hypergeometric distribution, it is found that there is a 0.3571 = 35.71% probability that he planted 2 daffodil bulbs and 4 tulip bulbs.
The bulbs are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.
Hypergeometric distribution:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- N is the size of the population.
- n is the size of the sample.
- k is the total number of desired outcomes.
In this problem:
- Total of 5 + 4 = 9 bulbs, hence [tex]N = 9[/tex].
- 5 are tulips, hence [tex]k = 5[/tex]
- Plants 6 bulbs, hence [tex]n = 6[/tex]
The probability that he planted 2 daffodil bulbs and 4 tulip bulbs is P(X = 4), hence:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 4) = h(4,9,6,) = \frac{C_{5,4}C_{4,2}}{C_{9,6}} = 0.3571[/tex]
0.3571 = 35.71% probability that he planted 2 daffodil bulbs and 4 tulip bulbs.
A similar problem is given at https://brainly.com/question/24826394
