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sasho [114]
2 years ago
10

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 the

ir next 15?
a. 0.6%
b. 21.4%
c. 19.8%
d. 11.1%​
Mathematics
2 answers:
Ghella [55]2 years ago
5 0

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)

Gekata [30.6K]2 years ago
3 0
  • 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

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