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We have three fair coins, each of which has probability 1/2 of having a heads outcome and a tails outcome. The experiment is to flip all three coins and observe the sequence of heads and tails. For example, outcome HTH means coin 1 was heads, coin 2 was tails, coin 3 was heads. Note that there are 8 total outcomes, and we assume that each one is equally likely. What is the probability that the outcome has at least two consecutive heads in the sequence?

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abidemiokin

Answer: 3/8

Step-by-step explanation:

Firstly, let's look at the possible outcome when 3 coins are tossed.

If two coins are first tossed, the possible outcome will be,

{HH, HT, TH, TT}

if one more coin is tossed together with the two to make it 3coins, the possible outcome will be gotten by matching the H and T of the third coin with the set of sample space above to give us,

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

This gives a total sample space of 8.

Outcomes that has at least two consecutive heads in the sequence are {HHH, HHT, THH} which is the possible outcome i.e 3

Probability that the outcome has at least two consecutive heads in the sequence will be;

Possible outcome/total outcome

= 3/8

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