Players a, b, and c toss a fair coin in order. the first to throw a head wins. what are their respective chances of winning

1 answer:

8 0

Probability a gets a head = 1/2
.. .. .. b .. .. ... .. = P(a gets tail) * 1/2 = 1/2 * 1/2 = 1/4
.. ... . .. c .. .. .. .. . = P(and and b gets tail) * 1/2 = 1/4 * 1/2 = 1/8

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Please help me anyone ???????

tester [92]

Probability =
(number or amount of possible successes)(total number or amount of possible outcomes)

Area of any circle is (pi) x (radius squared)

Area of the whole board = (pi) x (13 squared) = 169 pi square inches

Area of the shaded region = (pi) x (5 squared) = 25 pi square inches

Probability = (success area) / (total possible area) = 25pi/169pi = <u>0.15 </u>(rounded)

Notice that if you do it my way ... keep 'pi' in the fraction ... you don't even
need to know what the value of 'pi' is, because there's one in the top and
bottom of the fraction, and they cancel.

6 0

3 years ago

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

Oksana_A [137]

Answer:

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

${n \choose 5 } = \frac{n!}{5!(n-5)!}$

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 ${n \choose 4} .$

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is $4 * {n \choose 4} .$

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is ${n \choose 3} .$

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have ${3 \choose 2 } = 3$ possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of $6 * {n \choose 3}$

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of ${n \choose 2}$ possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.