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The probability is 0.6 that an “unfair” coin will turn up ta
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14 Mar 2019, 09:45
14 Mar 2019, 09:45
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The probability is 0.6 that an “unfair” coin will turn up tails on any given toss. If the coin is tossed 3 times, what is the probability that at least 1 of the tosses will turn up tails?
A. 0.064
B. 0.36
C. 0.64
D. 0.784
E. 0.936
A. 0.064
B. 0.36
C. 0.64
D. 0.784
E. 0.936
Re: The probability is 0.6 that an “unfair” coin will turn up ta
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20 Mar 2019, 15:27
20 Mar 2019, 15:27
3
Expert Reply
If at least 1 of the tosses is tails, then 1, 2, or all 3 tosses could be tails. The only possibility this leaves out is the possibility of no tails at all, or all three tosses being heads. So we can simply calculate that probability and subtract it from 1.
The probability of the first toss being heads is 0.4. The probability of the second toss being heads is also 0.4. Same for the third as well. Together, the probability of heads on the first AND second AND third tosses is \(0.4*0.4*0.4 = 0.064\) (generally speaking, in probability "and" means multiply).
So, the probability of this one specific case, all heads, not happening is \(1 - 0.064 = 0.936\).
The probability of the first toss being heads is 0.4. The probability of the second toss being heads is also 0.4. Same for the third as well. Together, the probability of heads on the first AND second AND third tosses is \(0.4*0.4*0.4 = 0.064\) (generally speaking, in probability "and" means multiply).
So, the probability of this one specific case, all heads, not happening is \(1 - 0.064 = 0.936\).
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Re: The probability is 0.6 that an “unfair” coin will turn up ta
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15 Jan 2020, 06:52
15 Jan 2020, 06:52
2
Carcass wrote:
The probability is 0.6 that an “unfair” coin will turn up tails on any given toss. If the coin is tossed 3 times, what is the probability that at least 1 of the tosses will turn up tails?
A. 0.064
B. 0.36
C. 0.64
D. 0.784
E. 0.936
A. 0.064
B. 0.36
C. 0.64
D. 0.784
E. 0.936
ASIDE---------------
We want P(select at least 1 tails)
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 tails) = 1 - P(not getting at least 1 tails)
What does it mean to not get at least 1 tails? It means getting zero tails.
So, we can write: P(getting at least 1 tails) = 1 - P(getting zero tails)
-------------------
P(getting zero tails)
P(getting zero tails) = P(heads on 1st toss AND heads on 2nd toss AND heads on 3rd toss)
= P(heads on 1st toss) x P(heads on 2nd toss) x P(heads on 3rd toss)
= 0.4 x 0.4 x 0.4
= 0.064
So, P(getting at least 1 tails) = 1 - 0.064 = 0.936
Answer: E
Cheers,
Brent
