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icang [17]
3 years ago
9

What numbers would be in the in the 4 sections of a 2 x 2 array that shows the problem 19 x 11?

Mathematics
1 answer:
marishachu [46]3 years ago
3 0

Answer:

number 2

Step-by-step explanation:

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You flip an unfairly weighted coin 4 times. If the probability of getting a tail is 0. 38.
anygoal [31]

Answer:

0.84306608

Step-by-step explanation:

The probability of an event, e, occurring exactly r times over n trials follows the formula

(n combination r) * p^r * q ^ (n-r)

with p being the probability the event will occur and q being the probability the event will not occur.

I assume you have a calculator/can find one online that can do combinations.

Here, we want to figure out if you get:

- 0 tails

- 1 tail

- 2 tails

If you get 3 or 4 tails, we are getting more than the 2 tails desired

For 0 tails:

- p is the probability a tail will occur = 0.38

- we want it to occur 0 times, so r = 0

- q is (1-p) = 1- 0.38 = 0.62

- we have 4 trials, as we flip it 4 times

(n combination r) * p^r * q ^ (n-r) = (4 combination 0) * (0.38) ^0 * (0.62) ^(4-0) = 1 * (0.38) ^0 * (0.62) ^(4-0) = 0.14776336

For 1 tail:

- p = 0.38, q = 0.62 as with 0 tails

- r = 1, n = 4

(n combination r) * p^r * q ^ (n-r) = (4 combination 1) * (0.38) ^1 * (0.62) ^(4-1) = 4* (0.38) ^1 * (0.62) ^(4-1) = 0.36225856

For 2 tails:

- p = 0.38, q = 0.62 as with 0 tails

- r = 2, n = 4

(n combination r) * p^r * q ^ (n-r) = (4 combination 2) * (0.38) ^1 * (0.62) ^(4-2) = 6* (0.38) ^2 * (0.62) ^(4-2) = 0.33304416

add our 3 probabilities together

0.33304416 + 0.36225856 + 0.14776336 = 0.84306608 as our answer

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