Q on binomial distribution probability

S

Sonal7

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I am not sure of the answers to part (b). In my opinion the answer should be b) 0.7^4 *0.3. a silly typing error right? Nice question with the part (c).
Screenshot 2020-08-07 at 20.59.53.png
 
"after the 4th" the answer given, is not the same as "on the 5th" which is what you recommended.

"after the 4th" can be 5th, 6th, 7th, etc... Lucas just has to miss the first four. Nothing after that matters -- well, except that we are assuming that he will, eventually, hit at least one, some day.
 
b) .7^4* .3 + .7^5*.3 + .7^6*.3 + ... = .7^4*.3 (.7^0 + .7^1+ .7^2 + ...)

= .7^4 * .3 (1/.3) = .7^4
 
tkhunny said:
"after the 4th" the answer given, is not the same as "on the 5th" which is what you recommended.

"after the 4th" can be 5th, 6th, 7th, etc... Lucas just has to miss the first four. Nothing after that matters -- well, except that we are assuming that he will, eventually, hit at least one, some day.
Click to expand...
Do you *see* that what happens after the 4th shot does not matter? Can you explain the logic to me. Thanks!
 
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