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s344n2d4d5 [400]
3 years ago
11

A Starburst candy package contains 12 individual candy pieces. Each piece is equally likely to be berry, orange, lemon, or cherr

y, independent of all other pieces. (a) What is the probability that a Starburst package has only berry or cherry pieces and zero orange or lemon pieces
Mathematics
1 answer:
Oksanka [162]3 years ago
4 0

Answer:

(a) Each piece is “berry or cherry” with probability p = 1/2. The proba-

bility of only berry or cherry pieces is p

12 = 1/4096.

Step-by-step explanation:

(b) Each piece is “not cherry” with probability 3/4. The probability all 12

pieces are “not pink” is (3/4)12 = 0.0317.

(c) For i = 1, 2, . . . , 6, let Ci denote the event that all 12 pieces are flavor i.

Since each piece is flavor i with probability 1/4, P[Ci

] = (1/4)12. Since

Ci and Cj are mutually exclusive,

P[F1] = P[C1 ∪ C2 ∪ · · · ∪ C4] = X

4

i=1

P[Ci

] = 4 P[C1] = (1/4)11

.

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we can rewritte our given equation as:

4sinxcosx-2cosx=0

By factoring 2cosx on the left hand side, we have

2cosx(2sinx-1)=0

This equation has 2 solutions when

\begin{gathered} cosx=0\text{ ...\lparen A\rparen} \\ and \\ 2sinx-1=0\text{ ...\lparen B\rparen} \end{gathered}

From equation (A), we obtain

x=\frac{\pi}{2}\text{ or }\frac{3\pi}{2}

and from equation (B), we have

\begin{gathered} sinx=\frac{1}{2} \\ which\text{ gives} \\ x=\frac{\pi}{6}\text{ or }\frac{5\pi}{6} \end{gathered}

On the other hand, we can find one more solution from the original equation by substituting x=0, that is,

\begin{gathered} 2ccos(2\times0)-2cos0=0 \\ which\text{ gives} \\ 2\times1-2\times1=0 \\ so\text{ 0=0} \end{gathered}

then, x=0 is another solution. In summary, we have obtained the following solutions:

\begin{gathered} x=0 \\ x=\frac{\pi}{2}\text{or}\frac{3\pi}{2}\text{ and } \\ x=\frac{\pi}{6}\text{or}\frac{5\pi}{6} \end{gathered}

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1/4, 8/15, 6/5
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