Answer :
Answer:
1/4 Chances.
0.25% Chances
Step-by-step explanation:
Two methods to answer the question.
Here are presented to show the advantage in using the product rule given above.
Method 1:Using the sample space
The sample space S of the experiment of tossing a coin twice is given by the tree diagram shown below
The first toss gives two possible outcomes: T or H ( in blue)
The second toss gives two possible outcomes: T or H (in red)
From the three diagrams, we can deduce the sample space S set as follows
S={(H,H),(H,T),(T,H),(T,T)}
with n(S)=4 where n(S) is the number of elements in the set S
tree diagram in tossing a coin twice
The event E : " tossing a coin twice and getting two tails " as a set is given by
E={(T,T)}
with n(E)=1 where n(E) is the number of elements in the set E
Use the classical probability formula to find P(E) as:
P(E)=n(E)n(S)=14
Method 2: Use the product rule of two independent event
Event E " tossing a coin twice and getting a tail in each toss " may be considered as two events
Event A " toss a coin once and get a tail " and event B "toss the coin a second time and get a tail "
with the probabilities of each event A and B given by
P(A)=12 and P(B)=12
Event E occurring may now be considered as events A and B occurring. Events A and B are independent and therefore the product rule may be used as follows
P(E)=P(A and B)=P(A∩B)=P(A)⋅P(B)=12⋅12=14
NOTE If you toss a coin a large number of times, the sample space will have a large number of elements and therefore method 2 is much more practical to use than method 1 where you have a large number of outcomes.
We now present more examples and questions on how the product rule of independent events is used to solve probability questions.
Answer:
0.25
Step-by-step explanation:
Probability is how likely something is to happen
The probability of an event happening = (number of ways it can happen) ÷ (total number of outcomes)
With a coin toss there are 2 outcomes: head or tail
The number of ways tossing a tail can happen is 1.
⇒ probability of getting a tail = 1/2 = 0.5
As any tosses are independent of any previous tosses, the probability of the second toss will not be affected by the first toss.
Therefore, the probability of getting a tail in the first toss = 0.5
and the probability of getting a tail in the second toss = 0.5
We can calculate the chances of two or more independent events by multiplying the chances.
Therefore, the probability of getting a tail in the first toss AND in the second toss = 0.5 x 0.5 = 0.25 (since they are independent events)