Question

A fair coin is tossed three times. find the probability of getting at most one head and two consecutive heads. plz answer ,plz answer ​

Answer 1

Total possible outcomes : {TTT},{TTH},{THT},{HTT},{THH},{HTH},{HHT},{HHH}. That totals to 8. The favorable outcomes (i.e., getting at least two heads) are {THH},{HTH},{HHT},{HHH} ; which totals to 4.

So, P(getting at least two heads) = No: of favorable outcomes/Total number of outcomes = 4/8 = 1/2 = 0.5

Answer 2

Answer:

The required probabilities are found to be as follows:

(i) P(at most one head) = 0.5

(ii) P(two consecutive heads) = 0.375

Step-by-step explanation:

The eight total outcomes which are possible in case three coins are tossed or a coin is tossed thrice are as follows:

{HHH}, {HHT}, {HTH}, {HTT}, {TTT}, {TTH}, {THT}, {THH}

Now, we have to find the probability of getting at most one head and two consecutive heads.

(i) Finding the probability of finding at most one head:

Number of possible outcomes where at most one head is obtained: 4

Probability of getting at most one head:

$P(at\ most\ one\ head)=\frac{4}{8}$

or we can say:

P(at most one head) = 0.5

(ii) Finding the probability of finding two consecutive heads:

Number of possible outcomes where two consecutive heads are obtained: 3

Probability of getting two consecutive heads:

$P(two\ consecutive\ heads)=\frac{3}{8}$

or we can say:

P(two consecutive heads) = 0.375

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