Taylor says ABCD and EFGH are congruent because he can map ABCD to EFGH by multiplying each side length by 1.5 and translating t
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The probability that, at the tip of the fourth round, each of the players has four coins is 5/192.
Given that game consists of 4 rounds and every round, four balls are placed in an urn one green, one red, and two white.
It amounts to filling in an exceedingly 4×4 matrix. Columns C₁-C₄ are random draws each round; row of every player.
Also, let be the quantity of nonzero elements in
Let .
Parity demands that and
must equal 2 or 4.
Case 1: =4 and
=4. There are
=3 ways to put 2-1's in
, so there are 3 ways.
Case 2: =2 and
=4. There are 3 ways to position the -1 in
, 2 ways to put the remaining -1 in
(just don't put it under the -1 on top of it!), and a pair of ways for one among the opposite two players to draw the green ball. (We know it's green because Bernardo drew the red one.) we are able to just double to hide the case of
for a complete of 24 ways.
Case 3: . There are 3 ways to put the -1 in
. Now, there are two cases on what happens next.
- The 1 in
goes directly under the -1 in
. There's obviously 1 way for that to happen. Then, there are 2 ways to permute the 2 pairs of 1,-1 in
and
. (Either the 1 comes first in
- The 1 in
Hence, there are 3(2+2×2)=18 ways for this case. There's a grand total of 45 ways for this to happen, together with 12³ total cases. The probability we're soliciting for is thus 45/(12³)=5/192
Hence, at the top of the fourth round, each of the players has four coins probability is 5/192.
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Answer:
(B) No. A binomial probability model applies to only two outcomes per trial.
Step-by-step explanation:
The binomial probability is the probability of having sucesses on
repeated trials of an experiment that can only have two outcomes. This is why it is called the binomial probability.
Since in our problem there are three possible outcomes, the binomial probability cannot be used.
The correct answer is (B)
(B) No. A binomial probability model applies to only two outcomes per trial.
bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.
le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.