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:

i think there was a typo. you wrote pat brings 3 quarts of ice-cream to the party. each serving of ice-cream is 1 6 quart. how many total servings of ice-cream does pat bring to the party? i think the bold is supposed to be 1/6. so 1 quart of ice cream = 6 servings. 6 * 3 = 18.