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# help me with this maths question! Watch

1. (Original post by SubAtomic)
No, the most likely outcome is 3 tails.

Am I getting troll'd? Night x
I'm afraid Mullah is correct. Her solution gave me an alternative answer that i have on my multiple choice. i initially chose 3 tailsbut it it incorrect. You are forgetting to multiply the P(HHT) and P(TTH) by 3.
2. (Original post by Student#123)
I'm afraid Mullah is correct. Her solution gave me an alternative answer that i have on my multiple choice. i initially chose 3 tailsbut it it incorrect. You are forgetting to multiply the P(HHT) and P(TTH) by 3.
It's late I have a good excuse I will convince myself I was wrong another time.
3. (Original post by Student#123)
ok. what formula did you use?

my formula is this

for second question:

0.3 probability delay for TL1 and 0.4 probability delay for TL2

therefore

0.7 probability no delay for TL1 and 0.6 probability no delay for TL2

here are all possible combination of total outcomes

delay - delay which the probability is (0.3*0.4)

delay - non delay which the probability is (0.3*0.6)

non delay - delay which the probability is (0.7*0.4)

non delay - non delay which the probability is (0.7*0.6)

so you add the probability of the bold as a fraction of probability of all possible outcome

total probability of 1 delay = (0.3*0.6)+(0.7*0.4)/(0.3*0.4)+(0.3*0.6)+(0.7*0.4)+(0.7*0.6)

which should being 0.46

4. (Original post by Mullah.S)
my formula is this

for second question:

0.3 probability delay for TL1 and 0.4 probability delay for TL2

therefore

0.7 probability no delay for TL1 and 0.6 probability no delay for TL2

here are all possible combination of total outcomes

delay - delay which the probability is (0.3*0.4)

delay - non delay which the probability is (0.3*0.6)

non delay - delay which the probability is (0.7*0.4)

non delay - non delay which the probability is (0.7*0.6)

so you add the probability of the bold as a fraction of probability of all possible outcome

total probability of 1 delay = (0.3*0.6)+(0.7*0.4)/(0.3*0.4)+(0.3*0.6)+(0.7*0.4)+(0.7*0.6)

which should being 0.46

that is incorrect.

P(not delay on 1st and not delay on 2nd)=0.7*0.6=0.42

so P(at least 1 delay)=1-0.42=0.58 which is correct.

thank you.
5. (Original post by Student#123)
that is incorrect.

P(not delay on 1st and not delay on 2nd)=0.7*0.6=0.42

so P(at least 1 delay)=1-0.42=0.58 which is correct.

thank you.

Oh i am think you say exactly only 1 delay,

not seeing the at least

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