Question

2 If a hockey team loses 33% of the games they play, what is the probability that this team will win exactly 10 games out of their next 15?

a. 0.6%
b. 21.4%
c. 19.8%
d. 11.1%​

Answer 1

• They lose 33% of games i.e 0.33
• They win 67% of games i e 0.67

So

• P(A)=0.33
• P(A')=0.67

Now

• They win from 15games=10
• They lose=5

The required probability

`\\ \rm\rightarrowtail P(A')^(10)* P(A)^5* ^(15)C_(10)`

`\\ \rm\rightarrowtail (0.33)^(5)* (0.67)^(10)* (15!)/(5!10!)`

`\\ \rm\rightarrow 0.0039(0.0182)*(15(14)(13)(12)(11)10!)/(5!10!)`

`\\ \rm\rightarrowtail 0.00007098* (360360)/(120)`

`\\ \rm\rightarrowtail 0.00007098(3003)`

`\\ \rm\rightarrowtail 0.213`

So probability

• 0.213(100)=21.3%

Option B

Answer 2

b. 21.4 % (nearest tenth)

Explanation:

P(lose) = 33% = 0.33

P(win) = 1 - 0.33 = 0.67

Using binomial distribution X ~ B(n, p)

where n is the number of trials and p is the probability of success

⇒ X ~ B(15, 0.67)

Binomial probability formula:

`\sf P(X=x)= \ _nC_x\cdot p^x \cdot(1-p)^(n-x)`

`\sf \implies P(X=10)=15C10\cdot 0.67^(10) \cdot(1-0.67)^(15-10)`

`\sf =3003 \cdot 0.67^(10) \cdot 0.33^5`

`\sf =0.214226434...`

Converting to percentage:

0.214266434... x 100% = 21.4 % (nearest tenth)