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Part 2: NO LINKS!! NOT MULTIPLE CHOICE! Please help me​

Part 2: NO LINKS!! NOT MULTIPLE CHOICE! Please help me​-example-1
User Adya
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5 votes

Answer:

(see attachment for tree diagram)


\sf Probability\:of\:an\:event\:occurring = (Number\:of\:ways\:it\:can\:occur)/(Total\:number\:of\:possible\:outcomes)

Part (a)


\textsf{P(Head) and P(3)}=\sf (1)/(2) * (1)/(6)=(1)/(12)

Part (b)


\textsf{P(Tail) and P(even)}=\sf (1)/(2) * (3)/(6)=(3)/(12)=(1)/(4)

Part (c)


\textsf{P(not 6)}=\sf 1-\textsf{P(6)}=1-(1)/(6)=(5)/(6)


\implies \textsf{P(Head) and P(not 6)}=\sf (1)/(2) * (5)/(6)=(5)/(12)

Part (d)

As the six-sided die does not have a side labelled "7", the probability of rolling a 7 is zero.


\implies \textsf{P(Head) and P(7)}=\sf (1)/(2) * 0=0

Part 2: NO LINKS!! NOT MULTIPLE CHOICE! Please help me​-example-1
User Pseyfert
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