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Two coins are tossed. Assume that each event is equally likely to occur. ​a) Use the counting principle to determine the number of sample points in the sample space. ​b) Construct a tree diagram and list the sample space. ​c) Determine the probability that no tails are tossed. ​d) Determine the probability that exactly one tail is tossed. ​e) Determine the probability that two tails are tossed. ​f) Determine the probability that at least one tail is tossed.

User Egor Neliuba
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1 Answer

9 votes
9 votes

Answer:

(a) 4 sample points

(b) See attachment for tree diagram

(c) The probability that no tail is appeared is 1/4

(d) The probability that exactly 1 tail is appeared is 1/2

(e) The probability that 2 tails are appeared is 1/4

(f) The probability that at least 1 tail appeared is 3/4

Explanation:

Given


Coins = 2

Solving (a): Counting principle to determine the number of sample points

We have:


Coin\ 1 = \{H,T\}


Coin\ 2 = \{H,T\}

To determine the sample space using counting principle, we simply pick one outcome in each coin. So, the sample space (S) is:


S = \{HH,HT,TH,TT\}

The number of sample points is:


n(S) = 4

Solving (b): The tree diagram

See attachment for tree diagram

From the tree diagram, the sample space is:


S = \{HH,HT,TH,TT\}

Solving (c): Probability that no tail is appeared

This implies that:


P(T = 0)

From the sample points, we have:


n(T = 0) = 1 --- i.e. 1 occurrence where no tail is appeared

So, the probability is:


P(T = 0) = (n(T = 0))/(n(S))

This gives:


P(T = 0) = (1)/(4)

Solving (d): Probability that exactly 1 tail is appeared

This implies that:


P(T = 1)

From the sample points, we have:


n(T = 1) = 2 --- i.e. 2 occurrences where exactly 1 tail appeared

So, the probability is:


P(T = 1) = (n(T = 1))/(n(S))

This gives:


P(T = 1) = (2)/(4)


P(T = 1) = (1)/(2)

Solving (e): Probability that 2 tails appeared

This implies that:


P(T = 2)

From the sample points, we have:


n(T = 2) = 1 --- i.e. 1 occurrences where 2 tails appeared

So, the probability is:


P(T = 2) = (n(T = 2))/(n(S))

This gives:


P(T = 2) = (1)/(4)

Solving (f): Probability that at least 1 tail appeared

This implies that:


P(T \ge 1)

In (c), we have:


P(T = 0) = (1)/(4)

Using the complement rule, we have:


P(T \ge 1) + P(T = 0) = 1

Rewrite as:


P(T \ge 1) = 1-P(T = 0)

Substitute known value


P(T \ge 1) = 1-(1)/(4)

Take LCM


P(T \ge 1) = (4-1)/(4)


P(T \ge 1) = (3)/(4)

Two coins are tossed. Assume that each event is equally likely to occur. ​a) Use the-example-1
User ConnectedSystems
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