Consider a biased coin where p({h}) = 1/3 and p({t}) = 2/3. we toss this biased coin 3 times and we observe the sequence of the heads and the tails. a) let a be the event that heads appear only in the third toss. find p(a). b) let b be the event that heads appear in the third toss. find p(b). c) let c be the event that heads appear two times. find p(c).

a) 4/27

b) 1/3

c) 6/27

Step-by-step explanation:

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a)

The event A has only one member:

T T H

Therefore:

P(A) = P(T)P(T)P(H) = (2/3)(2/3)(1/3) = 4/27

b)

For event B, we are only interested in the last toss

x x H

Since the first two tosses are not important for event B they do not contribute to P(B)

P(B) = 1 * 1* P(H) = 1/3

c)

The following are elements of C:

H H T

H T H

T H H

Therefore:

P(C) = P(HHT) + P(HTH) + P(THH)

We can easily see that the probability of the three members is the sam:

P(HHT) = (1/3) (1/3) (2/3) = 2/27

P(HTH) = (1/3) (2/3) (1/3) = 2/27

P(THH) = (2/3) (1/3) (2/3) = 2/27

Therefore:

P(C) = 3 P(HHT) = 6/27

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step-by-step explanation:

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.