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:

the inequality is:

250c+180g≥950

step-by-step explanation:

let c denotes the number of days in a week   miyoko worked as a cryptographer.

and g denotes the number of days in a week she works as a geologist.

now we are given that she earns $250 for every day she works as a cryptographer,and earns $180 for every day she works as a geologist.

now miyoko's goal is to earn more than $950 this week.

hence, the inequality could be represented as:

250c+180g≥950

susie has 7 apples