Sarah tosses 5 coins, one after the other.
a) How many different outcomes are possible?
b) Draw a tree diagram to illustrate the different possibilities.
c) In how many ways will the first coin turn up heads and the last coin turn up tails?
d) In how many ways will the second, third, and fourth coins all turn up tails?
e) Would the same possibilities arise in parts a), b), c), and d) if one coin was tossed 5 times in
succession?
a) To determine the number of different outcomes possible when tossing 5 coins, we can use the concept of the multiplication principle. Since each coin can have 2 possible outcomes (heads or tails), we can multiply the number of outcomes for each coin together.
Number of outcomes for 1 coin = 2 (heads or tails)
Number of outcomes for 5 coins = 2^5 = 32
Therefore, there are 32 different outcomes possible when tossing 5 coins.
b) To draw a tree diagram illustrating the different possibilities, we can start by listing the possible outcomes for each coin toss, and then branching out for each additional coin toss. Here's an example:
C1: H — T
| |
C2: H — T
| |
C3: H — T
| |
C4: H — T
| |
C5: H — T
c) To determine the number of ways the first coin will turn up heads and the last coin will turn up tails, we can break it down into individual events. The first coin has 2 possible outcomes (heads or tails), and the last coin also has 2 possible outcomes. Since these events are independent, we can multiply the number of outcomes for each event together.
Number of ways first coin turns up heads = 1 (since we want a specific outcome)
Number of ways last coin turns up tails = 1 (again, a specific outcome)
Number of ways first coin turns up heads and last coin turns up tails = 1 * 1 = 1
Therefore, there is 1 way for the first coin to turn up heads and the last coin to turn up tails.
d) To determine the number of ways the second, third, and fourth coins all turn up tails, we can apply the same logic as before. Each of these three coins has 2 possible outcomes (heads or tails), so we can multiply the number of outcomes for each coin together.
Number of ways second coin turns up tails = 1 (specific outcome)
Number of ways third coin turns up tails = 1 (specific outcome)
Number of ways fourth coin turns up tails = 1 (specific outcome)
Number of ways second, third, and fourth coins all turn up tails = 1 * 1 * 1 = 1
Therefore, there is 1 way for the second, third, and fourth coins to all turn up tails.
e) If one coin was tossed 5 times in succession, the possibilities would be different from the previous parts. In this case, each coin toss would only have 2 possible outcomes (heads or tails), but since we are tossing the same coin multiple times, the outcomes are not independent. Therefore, there would only be 2 outcomes for each toss (heads or tails), and the total number of different outcomes possible would be 2^5 = 32, which is the same as part a). However, the specific outcomes and combinations would be different.
a) 2^5
c) 2^3 ... arrangements of 3 middle coins
d) 2^2 ... arrangements of 2 end coins
e) yes , a coin toss is a coin toss