a coin will be tossed 10 times. Find the chance that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses

The chance that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses is P=0.0488.

Step-by-step explanation:

To solve this problem we divide the tossing in two: the first 5 tosses and the last 5 tosses.

Both heads and tails have an individual probability p=0.5.

Then, both group of five tosses have the same binomial distribution: n=5, p=0.5.

The probability that k heads are in the sample is:

Then, the probability that exactly 2 heads are among the first five tosses can be calculated as:

For the last five tosses, the probability that are exactly 4 heads is:

Then, the probability that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses can be calculated multypling the probabilities of these two independent events:

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."