For one toss of a certain coin, the probability

[SAM]

For one toss of a certain coin, the probability that the outcome is Heads is 0.6. If the coin is tossed 5 times , which of the following is the probability that the outcome will be heads atleast 4 times ?

The answer is 5(0.6)^4(0.4) + (0.6)^5. Though I was close and guessed (0.6)^4(0.4) + (0.6)^5, I am not sure why we multiply the first part with 5. Kindly suggest. Thx.

[RonPurewal]

For one toss of a certain coin, the probability that the outcome is Heads is 0.6. If the coin is tossed 5 times , which of the following is the probability that the outcome will be heads atleast 4 times ?

The answer is 5(0.6)^4(0.4) + (0.6)^5. Though I was close and guessed (0.6)^4(0.4) + (0.6)^5, I am not sure why we multiply the first part with 5. Kindly suggest. Thx.

first off, an admonishment: please follow the rules when posting questions.
your title should consist of the first 8 words (approximately) of the problem. since i am a true gentleman with a big heart, i have changed the title of this thread (and your other one) for you, but please title your posts correctly from now on.

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here's why:
if you take a single possibility that consists of 4 heads and 1 tail - such as HHHHT - the probability of that single possibility is (0.6)(0.6)(0.6)(0.6)(0.4), which is the left-hand term in your expression.
but there's the problem: there are 4 more ways in which the 4 heads/1 tail event can occur (namely HHHTH, HHTHH, HTHHH, THHHH). each of those carries the same probability as the above event (because all of them are four 0.6's and one 0.4, multiplied in some order); there are five possible versions of the event, total, hence the 5 in the answer.

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in general:
whenever you have Bernoulli trials, also called binomial trials - which means a fixed # of trials with the same probabilities of success and failure every time, like this example - the formula is: (number of ways to choose successes/failures)(p ^ # of successes)((1-p) ^ # of failures). the left-hand coefficient (the # of ways to choose) is a combination, which can be generated with our anagram method or with the traditional 'combinations' formula from algebra.
for instance, if you had to figure the probability of 3 successes out of 5, the front coefficient (in front of the probabilities) would be 5!/(2!3!) = 10.

[Justin]

Hi RPurewal,
I got how we got first part.
Can you tell me the RHS "+ (0.6)^5".
Or am I missing something

[RonPurewal]

Hi RPurewal,
I got how we got first part.
Can you tell me the RHS "+ (0.6)^5".
Or am I missing something

that's the probability of all heads.

there's only one way to do that, so you just multiply the probabilities:
(heads)(heads)(heads)(heads)(heads)
(0.6)(0.6)(0.6)(0.6)(0.6)
(0.6)^5

[Guest]

Ron, you last wrote to this thread:

in general:
whenever you have Bernoulli trials, also called binomial trials - which means a fixed # of trials with the same probabilities of success and failure every time, like this example - the formula is: (number of ways to choose successes/failures)(p ^ # of successes)((1-p) ^ # of failures). the left-hand coefficient (the # of ways to choose) is a combination, which can be generated with our anagram method or with the traditional 'combinations' formula from algebra.
for instance, if you had to figure the probability of 3 successes out of 5, the front coefficient (in front of the probabilities) would be 5!/(2!3!) = 10.

Can you provide more color on this formula - as it applies to the problem above, for example.

I want to make sure I understand the variables correctly.
(number of ways to choose successes/failures)(p ^ # of successes)((1-p) ^ # of failures)

for this problem, according to the formula above, the equation should look like this right?
1st
# of ways to choose successes = 5, derived as you stated from the 'combinations' formula 5 C 4, which = 5!/(3!2!) or 5
2nd
(p ^ # of successes)((1-p) ^ # of failures)
(.6)^4 x (.4)^1

So, if the question had asked:
For one toss of a certain coin, the probability that the outcome is Heads is 0.2. If the coin is tossed 10 times , which of the following is the probability that the outcome will be heads atleast 3 times?

The answer should be:
1st
10 C 3 = 10!/(7!3!) = 120
2nd
(.2)^3 x (.8)^7

together
HHHTTTTTTT as the single possibility and the 120 as the number of combinations of that possibility.
120 (.2)^3 x (.8)^7

then you would have to add that the combinations of 4H, 5H, 6H...and so on. Is that correct?


if so, my next question is for the " at least" part stated in the question - when do we know to use the OPPOSITE, (1-never) to solve the problem?
Clearly, adding each of those combinations would take way too long.

Thanks in advance.

[RonPurewal]

So, if the question had asked:
For one toss of a certain coin, the probability that the outcome is Heads is 0.2. If the coin is tossed 10 times , which of the following is the probability that the outcome will be heads atleast 3 times?

The answer should be:
1st
10 C 3 = 10!/(7!3!) = 120
2nd
(.2)^3 x (.8)^7

together
HHHTTTTTTT as the single possibility and the 120 as the number of combinations of that possibility.
120 (.2)^3 x (.8)^7

then you would have to add that the combinations of 4H, 5H, 6H...and so on. Is that correct?

absolutely correct.

if so, my next question is for the " at least" part stated in the question - when do we know to use the OPPOSITE, (1-never) to solve the problem?
Clearly, adding each of those combinations would take way too long.

Thanks in advance.

basically,
* consider all the cases that would be subsumed under the event itself
* consider all the cases that would be subsumed under the OPPOSITE event (i.e., the event not happening)
* see which of the 2 calculations would be less effort
* do that one

in most cases, the distinction will be pretty stark. for instance, if you flip 8 times and you have to calculate the probability of at least 2 heads, then there's no question: the direct calculation would be 2h + 3h + 4h + 5h + 6h + 7h + 8h, but the indirect calculation is just 1 - (0h + 1h).

also, you'd do well to learn to recognize the signal words - such as "at least" - that clue you in to the possibility of the indirect approach. "at least" isn't the only signal; basically, any signal that indicates the potential for a large number of possibilities should alert you to the potential for the indirect (1 - opposite) approach.