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Answer:
The answer is 0.008789.
Step-by-step explanation:
The probability of getting a head or tail in a toss is [tex]\frac{1}{2}[/tex].
It is given that in the 6th toss, the outcome should be head.
Rather than the 6th toss, there are 9 tosses. From these 9 tosses 1 toss can be selected in 9 ways.
Hence, the probability is [tex]9\times\frac{1}{2^{10}}[/tex] = Â 0.008789.
The probability at 6th coin tossed out of 10 independent coin tosses comes as head given that exactly 2 heads are there is 0.2
How to calculate the probability of an event?
Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.
Then, suppose we want to find the probability of an event E.
Then, its probability is given as
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}}[/tex]
where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.
In the given condition, we see that 6th coin can either belong to that 2 heads group or not. If it belongs to that 2 heads group, then it has got head, else it has got tail.
Thus, we get:
P(Getting head on 6th coin given that exactly 2 coins out of 10 independent coins tossed are head) = P(6th coin belongs to those 2 coins who got head)
The probability that the 6th coin belongs to those 2 heads output coin is 2/10 (since there are 2 favorable cases, and 10 total cases). This is the exact probability of that 6th coin getting head.
Thus, P(Getting head on 6th coin | exactly 2 heads are there in 10 tossed coins) = 2/10 = 0.2
where we have: P(A|B) = probability of occurrence of A given B has already occurred
Thus, The probability at 6th coin tossed out of 10 independent coin tosses comes as head given that exactly 2 heads are there is 0.2
Learn more about conditional probability here:
https://brainly.com/question/10739997
