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Vladimir79 [104]
3 years ago
6

Jason and Kyle both choose a number from 1 to 10 at random. What is the probability that both numbers are odd?

Mathematics
1 answer:
Bezzdna [24]3 years ago
5 0
Half of the numbers between 1 to 10 are odd, 1,3,5,7,9.
So the probability is 1/2. 
Hope this helps. 
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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
tankabanditka [31]
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
6 0
3 years ago
Read 2 more answers
-2=3(y-2) how do you solve this problem for y​
Ksivusya [100]

Answer:

y= 4/3

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

−2=3(y−2)

−2=(3)(y)+(3)(−2)(Distribute)

−2=3y+−6

−2=3y−6

Step 2: Flip the equation.

3y−6=−2

Step 3: Add 6 to both sides.

3y−6+6=−2+6

3y=4

Step 4: Divide both sides by 3.

3y /3 = 4 /3

y= 4/3

Answer:

y= 4/3

8 0
3 years ago
In which two of these numbers does the digit
monitta

Answer:

85.62

Step-by-step explanation:

Hope this helps!

If not, I am sorry.

7 0
2 years ago
Read 2 more answers
How much does he/she pay with 10 %
ss7ja [257]

Answer:

5085

Step-by-step explanation:

price after discount is 5000- 10% of 5000 = 4500

V.A.T = 13% of 4500 = (13/100)*4500= 13*45= 585

Price after V.A.T = 4500 + 585 = 5085

5 0
3 years ago
Help with this problem please!!!!!
mariarad [96]

Answer:

3

Step-by-step explanation:

3 0
3 years ago
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