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3 years ago

When two fair dice are rolled, what is the probability that at least one of the numbers will be even??

1 answer:

4 0

The probability would be 1/4. The probability of rolling an even number on each die is 1/2. Since there are two dice you would multiply the probability of each die together. 1/2*1/2=1/4. The reason you would this is because you could roll two odd numbers that equal an even number such as 3 and 5.

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No spams<br>easy one , thx for answering ~

SSSSS [86.1K]

Answer:

Step-by-step explanation:

l// m , t is transversal.

Corresponding angles are equal.

3x - 20 = 2x + 10

Add 20 to both sides

3x = 2x + 10 + 20

3x = 2x + 30

Subtract 2x from both sides

3x - 2x = 30

x = 30

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7/18 ratio in simplest form

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How many times can 49 go into 817

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3 years ago

Read 2 more answers

A simplified model for the movement of the price of a stock supposes that on each day the stock’s price either moves up 1 unit w

DanielleElmas [232]

Answer:

(a) The probability that after 2 days the stock will be at its original price is

$P_a=2p(1-p)$

(b) The probability that after 3 days the stock’s price will have increased by 1 unit is

$P_b=3p^2(1-p)$

(c) Given that after 3 days the stock’s price has increased by 1 unit, the probability that it went up on the first day is

$P=2/3$

Step-by-step explanation:

(a) What is the probability that after 2 days the stock will be at its original price?

For the price of the stock to be in its original price, there are two ways it can happen:

1) the price increases the first day by one unit and decreases by one unit the second day. The probability of this event is P=p*(1-p).

2) the price decreases the first day by one unit and increases by one unit the second day. The probability of this event is P=(1-p)p.

The probability that after 2 days the stock will be at its original price is the sum of the probability of this two events:

$P=P_1+P_2=p(1-p)+(1-p)p=2p(1-p)$

(b) What is the probability that after 3 days the stock’s price will have increased by 1 unit?

For this event to happen (one unit increase in 3 days), it must have happened 2 increases in price and 1 decrease.

There are 3 possible ways of this to happen:

1) decrease in the first day.

2) decrease in the second day.

3) decrease in the third day.

Each one ot this 3 events has the same probability $P_i=p^2(1-p)$ .

So the probability that after 3 days the stock’s price will have increased by 1 unit is equal to the sum of the probabilities of this events:

$P=P_1+P_2+P_3=3p^2(1-p)$

(c) Given that after 3 days the stock’s price has increased by 1 unit, what is the probability that it went up on the first day?

According to the answer (c), there are 3 events where the price has increase by one unit after 3 days, each one with equal probability $P_i=p^2(1-p)$ .

Of these 3 events, there are 2 that have an increase in the first day. So we can conclude that, if the events have the same probability, the probability of the increase in the first day, given that the price has increased by one unit in 3 days, is P=2/3.

8 0

4 years ago

In a closed box there are 7 blue balls and 5 red balls. Randomly pick one from the box one at a time until the ball is green, th

timofeeve [1]

Answer:

1/7 is the probabiltiy

Step-by-step explanation:

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3 years ago