Using the principle of probability, the probability of the events in the question are :
- <em>Sample space = (H1, H2, H3, H4, H5, H6, THT, TTT, HTH, HTT) </em>
<u>To obtain the sample space, take the following steps</u> :
<em>Coin toss = ( H, T) </em>
<em>Die throw = (1, 2, 3, 4, 5, 6)</em>
<u>First experiment</u> :
<u>If H : roll a die once :</u>
<u>If T ; toss coin twice :</u>
- First toss : (TH, TT)
- Second toss : (THT, TTT, HTH, HTT)
A.) The ten elements of the sample space :
- (H1, H2, H3, H4, H5, H6, THT, TTT, HTH, HTT)
B.)
<u>Event A = exactly one head occurs </u>
P(A) = required outcome / Total possible outcomes
P(A) = 8/10 = 4/5
C.)
<u>Event B = atleast two tails or a number greater than 4 </u>
atleast 2 tails or number greater than 4 = 5
P(B) = 5/10 = 1/2
Therefore, the probability of event B is 0.5.
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