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3 years ago

Prove that the conjecture is false by providing a counter example.

Mathematics

1 answer:

Bas_tet [7]3 years ago

6 0

Answer:

Any negative number would be a counterexample.

Step-by-step explanation:

Since negative numbers get smaller as you multiply them by a larger number, this would prove this conjecture false.

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Suppose you pay a dollar to roll two dice. if you roll 5 or a 6 you Get your dollar back +2 more just like it the goal will be t

LiRa [457]

Answer:

(a)$67

(b)You are expected to win 56 Times

(c)You are expected to lose 44 Times

Step-by-step explanation:

The sample space for the event of rolling two dice is presented below

$(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)\\(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)\\(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)\\(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)$

Total number of outcomes =36

The event of rolling a 5 or a 6 are:

$(5,1), (6,1)\\ (5,2), (6,2)\\( (5,3), (6,3)\\ (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)$

Number of outcomes =20

Therefore:

P(rolling a 5 or a 6) $=\dfrac{20}{36}$

The probability distribution of this event is given as follows.

$\left|\begin{array}{c|c|c}$Amount Won(x)&-\$1&\$2\\&\\P(x)&\dfrac{16}{36}&\dfrac{20}{36}\end{array}\right|$

First, we determine the expected Value of this event.

Expected Value

$=(-\$1\times \frac{16}{36})+ (\$2\times \frac{20}{36})\\=\$0.67$

Therefore, if the game is played 100 times,

Expected Profit =$0.67 X 100 =$67

If you play the game 100 times, you can expect to win $67.

(b)

Probability of Winning $=\dfrac{20}{36}$

If the game is played 100 times

Number of times expected to win

$=\dfrac{20}{36} \times 100\\=56$ times$

Therefore, number of times expected to loose

= 100-56

=44 times

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