I need help with math homework please! Just two questions! Really don't get it ...?

Question

1. Four coins are tossed. What is the probability that they will show 3 heads and a tail?
the answer that the teacher gave to this one is 1/4.

2. Janet has 3 dimes and 6 nickels in her pocket. She selects 3 coins without replacement. What is the probability that Janet selects at least 2 nickels?
the answer that the teacher gave to this one is 65/84.

i tried several different ways to solve them, but no luck.

Answer 1

You can use the Pascal triangle for this one:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

and so forth. you get the elements of a row by adding pairs of elements from the previous row, and starting and ending the row with 1s. Look at the last row. The first 1 is the number of ways that, say, 4 heads show up. The 4 is the number of ways that 3 heads and one tail show up. The 6 is the number of ways that you get 2 heads and 2 tails and so forth. The total of all the ways is 16. So the probability of getting 3 heads and a tail is 4/16 = 1/4.

There is a formula for this, too. It is

C(n,r) = n!/r!(n-r)! = 4!/1!3! = 4/1 (the rest cancel out) = 4

2. If you draw 3 coins in a row, and two are nickels, they could come up DNN or NDN or NND or NNN. Compute the probability of each.

DNN. (3/9)*(6/8)*(5/7) = 5/28
Why? because the first coin has 3 chances out of 9 of being a dime, the second 6 chances out of 8 of being a nickel, now that a dime is gone, and the third 5 chances out of 7 of being a nickel.

NDN. (6/9)*(3/8)*(5/7) = 5/28

NND. (6/9)*(5/8)*(3/7) = 5/28

NNN. (6/9)*(5/8)*(4/7) = 5/21

Now add these together, and the result is 65/84.

Answer 2

1.) The best way to solve this is to list the possible solutions that would satisfy the conditions set. Because there are only 4 coins, this is possible to do. You can have:

H H H T
H H T H
H T H H
T H H H

That's 4 ways to get 3 heads and 1 tail. Now, how many possible answers are there in the whole universe?

Well, each coin can have two answers, heads or tails. So the total number of possible answers is:

2x2x2x2, or 2^4, which is 16.

Divide your number of ways to have 3 heads and 1 tail by the total number of possible answers, and you get 4/16, or 1/4.

Answer 3

1. Let the result of each individual toss be denoted "H" or "T". There are 16 different results: HHHH, HHHT, HHTH, etc. There are 4 ways to get 3 H's and 1 T: HHHT, HHTH, HTHH, THHH. 4 ways out of 16 = 1/4.

2. There are 9C3 = 9!/(6!3!) = 84 ways to select 3 coins w/o replacement. There is only 1 way to select 0 nickels (3 dimes). How many ways are there to select 1 nickel (2 dimes)? 3C2 ways to select 2 dimes * 6C1 ways to select 1 nickel = 3*6 = 18. Therefore, there are 19 ways to select less than 2 nickels, so there are 84-19 = 65 ways to select 2 or more nickels ==> 65/84.

Answer 4

1. Four coins are tossed. What is the probability that they will show 3 heads and a tail?

p(head) = 1/2
p(tail) = 1/2

p(3h and 1t) = (4C3)(1/2)³(1/2) = 4(1/8)(1/2) = 4/16 = 1/4

The idea is there are four different ways to get 3 heads and a tail. The tail can be the first, second, third or fourth coin.

Answer 5

The second question: when you pull one coin out the chance for a nickel to be pulled out is six times out of nine possibilities. The second try will be 5 times out of eight possibilities. So
6/9 times 5/8 equals 30/72 or 15/36 not the solution method in which the other guy attempted.
But since it takes a third try the chance would be subtracting 4 out of seven posibilites....

Thus 15/36 times 4/7= 60/252 or 30/126 or 15/58 No luck for me either!

Answer 6

right now i can only think of the first question.

the chance of tossing 2 heads and 2 tails is 1/2, if you were to take one of those tails and turn it into a head, it is also 1/2. Multiplying to two probabilities together will give you 1/4.

I hope someone else can explain the second one.

Answer 7

1st question: 1/4 is right, because probability is to get 1 heads and one tails when 2 coins are flipped, so take them out of the equation. New Equation: probability to flip 2 coins and get 2 heads; 1/2 times 1/2. that equals 1/4. your teacher is right.

sorry, can't get the 2nd one rite now.

Answer 8

A is an trouble-free substitution. In any subject including this you write the unique and the equalities first, merely as you probably did. 4x^2 + 5y + 7, whilst x=2 and y=5 as an occasion then you definately replace the constants for the variables, and instruct them in parentheses. 4(2)^2 + 5(5) + 7 then you definately simplify as many situations as obtainable. 4(4) + 25 + 7 sixteen + 25 + 7 40-one + 7 40 8 you're completed! For fixing for the a number of variables, you desire to get that variable by skill of itself on one facet of the equivalent sign. to try this, opposite the operations, one after the different, and do the comparable element to the two aspects. So, if the splendid facet is split by skill of two, you multiply the two aspects by skill of two, which in fact strikes the two from being divided on the splendid to being accelerated on the left. (the two's on the splendid will cancel out.) This additionally works for variables. If t is further, subtract T from the two aspects. on account which you do no longer understand what t is, it may ensue as '-t'. sturdy luck!