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sdas [7]
3 years ago
5

A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d

rag the sentences to match probabilities.
Possible probabilities are:
0.07
0.12
0.44
0.49


Options for answers are:
The probability that a randomly chosen team includes all 6 girls in the class.
The probability that a randomly chosen team has 3 girls and 7 boys.
The probability that a randomly chosen team has either 4 or 6 boys.
The probability that a randomly chosen team has 5 girls and 5 boys.
Mathematics
2 answers:
tankabanditka [31]3 years ago
6 0
The selection of r objects out of n is done in

C(n, r)= \frac{n!}{r!(n-r)!} many ways.

The total number of selections 10 that we can make from 6+7=13 students is 

C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}=  286
thus, the sample space of the experiment is 286

A. 
<span>"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.


</span>C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35
<span>
P(all 6 girls chosen)=35/286=0.12

B.
"</span>The probability that a randomly chosen team has 3 girls and 7 boys.<span>"

with the same logic as in A, the number of groups were all 7 boys are in, is 

</span>C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20
<span>
so the probability is 20/286=0.07

C.
"</span>The probability that a randomly chosen team has either 4 or 6 boys.<span>"

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is </span>0.12.
<span>
case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls. 

</span>C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105
<span>
the probability is 105/286=0.367

since  case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"</span><span>The probability that a randomly chosen team has 5 girls and 5 boys.</span><span>"

selecting 5 boys and 5 girls can be done in 

</span>C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126

many ways,

so the probability is 126/286=0.44
3241004551 [841]3 years ago
3 0

Answer:

 0.07 -The probability that a randomly chosen team has 3 girls and 7 boys.

0.12 -The probability that a randomly chosen team includes all 6 girls in the class.

0.44 -The probability that a randomly chosen team has 5 girls and 5 boys.

0.49-The probability that a randomly chosen team has either 4 or 6 boys.

GANGGGGG  #Plutolivesmatter

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Step-by-step explanation:

According to this question, the mean of 100 observation is found to be 40. This means that the number of values in the set of data is 100. However, at the time of computation two items were wrongly taken as 30 and 27 instead of 3 and 72.

The mean is derived by dividing the sum of values in the data (Σx) by the number of observations (n) i.e.

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