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

$\\ \rm\rightarrowtail P(A')^{10}\times P(A)^5\times ^{15}C_{10}$

$\\ \rm\rightarrowtail (0.33)^{5}\times (0.67)^{10}\times \dfrac{15!}{5!10!}$

$\\ \rm\rightarrow 0.0039(0.0182)\times\dfrac{15(14)(13)(12)(11)10!}{5!10!}$

$\\ \rm\rightarrowtail 0.00007098\times \dfrac{360360}{120}$

$\\ \rm\rightarrowtail 0.00007098(3003)$

$\\ \rm\rightarrowtail 0.213$

So probability

• 0.213(100)=21.3%

Option B

Answer 2

Answer:

b.  21.4 % (nearest tenth)

Step-by-step 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)