How many times does 44 go into 6,668??? Plz

A friend who works in a big city owns two cars, one small and one large. One-quarter of the time he drives the small car to work

lubasha [3.4K]

Answer:

The probability that he drove the small car is 0.318.

Step-by-step explanation:

We are given that a friend who works in a big city owns two cars, one small and one large. One-quarter of the time he drives the small car to work, and three-quarters of the time he takes the large car.

If he takes the small car, he usually has little trouble parking and so is at work on time with probability 0.7. If he takes the large car, he is on time to work with probability 0.5.

Let the Probability that he drives the small car = P(S) = $\frac{1}{4}$ = 0.25

Probability that he drives the large car = P(L) = $\frac{3}{4}$ = 0.75

Also, let WT = event that he is at work on time

So, Probability that he is at work on time given that he takes the small car = P(WT / S) = 0.7

Probability that he is at work on time given that he takes the large car = P(WT / L) = 0.5

Now, given that he was at work on time on a particular morning, the probability that he drove the small car is given by = P(S / WT)

We will use the concept of Bayes' Theorem for calculating above probability;

So, P(S / WT) = $\frac{P(S) \times P(WT/S)}{P(S) \times P(WT/S)+P(L) \times P(WT/L)}$

= $\frac{0.25 \times 0.7}{0.25 \times 0.7+0.75 \times 0.5}$

= $\frac{0.175}{0.55}$

= 0.318

Hence, the required probability is 0.318.

8 0

3 years ago

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

Compare 5.7145... and √29

myrzilka [38]

5.7145 is bigger than the square root of 29 (which is 5.3851...)

3 0

4 years ago

Rewrite the expression below as the sum of one constant and one variable term 2(x-4)+6x-5x•3

Lorico [155]

Answer:

-7x-8

Step-by-step explanation: