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hodyreva [135]
2 years ago
9

A box contains 3 coins, one fair, one weighted with a 2 3 chance of coming up heads, and one weighted with a 2 3 chance of comin

g up tails. You take a coin at random from the box, flip it 20 times, and get 11 heads. What is the probability you have the fair coin?
Mathematics
1 answer:
ddd [48]2 years ago
7 0

Answer:

The probability that we have a fair coin is 0.2717

Step-by-step explanation:

Let Ci be the event that the coin i is used for while we let i = 1,2,3 (representing the three coin)

We also let H be the event that the coin which is flipped lands (which is head)

Therefore, the problem arise:

That:  P(H/C1) = 1/2.............fair coin

          p(H/C2) = 2/3..........Chance of head

          p(H/C3) = 2/3............Chance of tail

Now, noting that the coin was picked at random,

We have, : p(Ci) = P(C1) = P(C2) = P(C3) = 1/3

We can then say that or calculate thus:

P( C2 | H ) = P (H ∩ C2) ÷ P(H)

Where P (H ∩ C2) means the probability of event intersection

= P(H ∩ C2) ÷ P(H ∩ C1) + P(H ∩ C2) + P(H ∩ C3)

= P(H | C2) P(C2) ÷ P(H | C1) P(C1) + P(H | C2) P(C2) + P(H | C3) P(C3)

= (1 / 2) (1 / 3) ÷  (1/2)(1/3) + (2/3)(1/3) + (2/3) (1/3)

0.5 × 0.3 ÷ (0.5 × 0.3) + (0.67 × 0.3) + (0.67 × 0.3)

0.15 ÷ 0.15 + 0.201 + 0.201

=0.15 / 0.552

= 0.2717

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Jenna has at most $80 to spend on clothes. She wants to buy a pair of jeans for$22 and spend the rest on t-shirts. Each t-shirt
weqwewe [10]

Answer:

7 shirts

Step-by-step explanation:

You first have to subtract 22 from 80, since that's how much she already spent, 80-22=58. Now you have to do 58/8, but since 8 doesn't divide evenly into 58, you have to go with the closet multiple of 8 which would be 56. 7x8=56, so Jenna can buy a maximum number of 7 shirts.

5 0
2 years ago
A sandbox has an area of 10 2/3 square feet, and the length is 1/4 feet. What is the width of the sandbox?
AlekseyPX

Answer:

The width of the sandbox is 42\frac{2}{3} \ ft.

Step-by-step explanation:

Given,

Area of the sandbox = 10\frac{2}{3}=\frac{32}{3}\ ft^2

Length = \frac{1}{4}\ ft

Solution,

Let the width of the sandbox be 'w'.

Since the sandbox is in shape of rectangle.

So we use the formula of area of rectangle.

Area = Length\times Width

On substituting the given values, we get;

\frac{1}{4}\times w=\frac{32}{3}\\

By cross multiplication method, we get;

w=\frac{32\times4}{3\times 1} =\frac{128}{3} =42\frac{2}{3} \ ft

Hence The width of the sandbox is 42\frac{2}{3} \ ft.

5 0
3 years ago
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