Coin Tosses Probability
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Naveen's coin box contains 8 fair standard coins (heads and tails) and 1 coin which has heads on both sides. He selects a coin randomly and flips it 4 times, getting all heads. If he flips this coin again, what is the probability it will be heads?
My solution:
Coin tosses are independent.
P(Head) = P(Head/Fair Coin)P(Fair Coin) + P(Head/Unfair Coin)P(Unfair Coin)
= 1/2 * 8/9 + 1 * 1/9
= 4/9 + 1/9
=5/9
Am I right?
probability conditional-probability
asked Jan 24 at 10:50
Beyond.MultiverseBeyond.Multiverse
114
$endgroup$
add a comment |
$begingroup$
Naveen's coin box contains 8 fair standard coins (heads and tails) and 1 coin which has heads on both sides. He selects a coin randomly and flips it 4 times, getting all heads. If he flips this coin again, what is the probability it will be heads?
My solution:
Coin tosses are independent.
P(Head) = P(Head/Fair Coin)P(Fair Coin) + P(Head/Unfair Coin)P(Unfair Coin)
= 1/2 * 8/9 + 1 * 1/9
= 4/9 + 1/9
=5/9
Am I right?
probability conditional-probability
asked Jan 24 at 10:50
Beyond.MultiverseBeyond.Multiverse
114
$endgroup$
add a comment |
$begingroup$
Naveen's coin box contains 8 fair standard coins (heads and tails) and 1 coin which has heads on both sides. He selects a coin randomly and flips it 4 times, getting all heads. If he flips this coin again, what is the probability it will be heads?
My solution:
Coin tosses are independent.
P(Head) = P(Head/Fair Coin)P(Fair Coin) + P(Head/Unfair Coin)P(Unfair Coin)
= 1/2 * 8/9 + 1 * 1/9
= 4/9 + 1/9
=5/9
Am I right?
probability conditional-probability
asked Jan 24 at 10:50
Beyond.MultiverseBeyond.Multiverse
114
$endgroup$
Naveen's coin box contains 8 fair standard coins (heads and tails) and 1 coin which has heads on both sides. He selects a coin randomly and flips it 4 times, getting all heads. If he flips this coin again, what is the probability it will be heads?
My solution:
Coin tosses are independent.
P(Head) = P(Head/Fair Coin)P(Fair Coin) + P(Head/Unfair Coin)P(Unfair Coin)
= 1/2 * 8/9 + 1 * 1/9
= 4/9 + 1/9
=5/9
Am I right?
probability conditional-probability
probability conditional-probability
asked Jan 24 at 10:50
Beyond.MultiverseBeyond.Multiverse
114
asked Jan 24 at 10:50
Beyond.MultiverseBeyond.Multiverse
114
asked Jan 24 at 10:50
Beyond.MultiverseBeyond.Multiverse
114
asked Jan 24 at 10:50
Beyond.MultiverseBeyond.Multiverse
114
asked Jan 24 at 10:50
Beyond.MultiverseBeyond.Multiverse
114
114
add a comment |
add a comment |
2 Answers
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$begingroup$
You have not made use of the fact that the coin he picked came up heads 4 times. You need to introduce an extra term in each summand of your computation for the chance that this kind of coin will come up heads four times in a row.
answered Jan 24 at 10:53
quaraguequarague
400110
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add a comment |
$begingroup$
Guide.
Let $H_i$ denote the event that the $i$-th throw results in a head.
Let $E$ denote the event that a fair coin was selected.
To be found is $P(H_5mid bigcap_{i=1}^4H_i)$ which can be done on base of:
$P(H_5mid bigcap_{i=1}^4H_i)=P(bigcap_{i=1}^5H_i)/P(bigcap_{i=1}^4H_i)$
$P(bigcap_{i=1}^nH_i)=P(bigcap_{i=1}^nH_imid E)P(E)+(bigcap_{i=1}^nH_imid E^{complement})P(E^{complement})$ for every $n$.
answered Jan 24 at 11:03
drhabdrhab
103k545136
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2 Answers
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oldest
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2 Answers
2
active
oldest
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$begingroup$
You have not made use of the fact that the coin he picked came up heads 4 times. You need to introduce an extra term in each summand of your computation for the chance that this kind of coin will come up heads four times in a row.
answered Jan 24 at 10:53
quaraguequarague
400110
$endgroup$
add a comment |
$begingroup$
You have not made use of the fact that the coin he picked came up heads 4 times. You need to introduce an extra term in each summand of your computation for the chance that this kind of coin will come up heads four times in a row.
answered Jan 24 at 10:53
quaraguequarague
400110
$endgroup$
add a comment |
$begingroup$
You have not made use of the fact that the coin he picked came up heads 4 times. You need to introduce an extra term in each summand of your computation for the chance that this kind of coin will come up heads four times in a row.
answered Jan 24 at 10:53
quaraguequarague
400110
$endgroup$
You have not made use of the fact that the coin he picked came up heads 4 times. You need to introduce an extra term in each summand of your computation for the chance that this kind of coin will come up heads four times in a row.
answered Jan 24 at 10:53
quaraguequarague
400110
edited Jan 24 at 11:47
answered Jan 24 at 10:53
quaraguequarague
400110
answered Jan 24 at 10:53
quaraguequarague
400110
answered Jan 24 at 10:53
quaraguequarague
400110
400110
add a comment |
add a comment |
$begingroup$
Guide.
Let $H_i$ denote the event that the $i$-th throw results in a head.
Let $E$ denote the event that a fair coin was selected.
To be found is $P(H_5mid bigcap_{i=1}^4H_i)$ which can be done on base of:
$P(H_5mid bigcap_{i=1}^4H_i)=P(bigcap_{i=1}^5H_i)/P(bigcap_{i=1}^4H_i)$
$P(bigcap_{i=1}^nH_i)=P(bigcap_{i=1}^nH_imid E)P(E)+(bigcap_{i=1}^nH_imid E^{complement})P(E^{complement})$ for every $n$.
answered Jan 24 at 11:03
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Guide.
Let $H_i$ denote the event that the $i$-th throw results in a head.
Let $E$ denote the event that a fair coin was selected.
To be found is $P(H_5mid bigcap_{i=1}^4H_i)$ which can be done on base of:
$P(H_5mid bigcap_{i=1}^4H_i)=P(bigcap_{i=1}^5H_i)/P(bigcap_{i=1}^4H_i)$
$P(bigcap_{i=1}^nH_i)=P(bigcap_{i=1}^nH_imid E)P(E)+(bigcap_{i=1}^nH_imid E^{complement})P(E^{complement})$ for every $n$.
answered Jan 24 at 11:03
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Guide.
Let $H_i$ denote the event that the $i$-th throw results in a head.
Let $E$ denote the event that a fair coin was selected.
To be found is $P(H_5mid bigcap_{i=1}^4H_i)$ which can be done on base of:
$P(H_5mid bigcap_{i=1}^4H_i)=P(bigcap_{i=1}^5H_i)/P(bigcap_{i=1}^4H_i)$
$P(bigcap_{i=1}^nH_i)=P(bigcap_{i=1}^nH_imid E)P(E)+(bigcap_{i=1}^nH_imid E^{complement})P(E^{complement})$ for every $n$.
answered Jan 24 at 11:03
drhabdrhab
103k545136
$endgroup$
Guide.
Let $H_i$ denote the event that the $i$-th throw results in a head.
Let $E$ denote the event that a fair coin was selected.
To be found is $P(H_5mid bigcap_{i=1}^4H_i)$ which can be done on base of:
$P(H_5mid bigcap_{i=1}^4H_i)=P(bigcap_{i=1}^5H_i)/P(bigcap_{i=1}^4H_i)$
$P(bigcap_{i=1}^nH_i)=P(bigcap_{i=1}^nH_imid E)P(E)+(bigcap_{i=1}^nH_imid E^{complement})P(E^{complement})$ for every $n$.
answered Jan 24 at 11:03
drhabdrhab
103k545136
answered Jan 24 at 11:03
drhabdrhab
103k545136
answered Jan 24 at 11:03
drhabdrhab
103k545136
answered Jan 24 at 11:03
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
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