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All questions? oe just the last two?Could anyone help me with these problems?Would be greatly appreciatedView attachment 59298
And please upload a clear picture
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All questions? oe just the last two?Could anyone help me with these problems?Would be greatly appreciatedView attachment 59298
Can you explain the "cross product....."again.
Can you explain the "cross product....."again.
Thanks

The binomial distribution is only applicable, when the probability of success remains constant. As you can see here, this is not the case, as the probability of success changes as the wrapped thingy is not replaced.View attachment 59309
s12-62
Part v
Ok so solving it using the combination probability formula works
but then i did it this way and the answer is wrong... pls tell why it can't work this way??
P(wrapped in gold foil) = 12/30 = 0.4
therefore , P(success) = 0.4 and P(failure) = 0.6
P(exactly 2 wrapped) = 4C2*0.4^2*0.6^2
= 0.346
But the answer is 0.368 which I know how to get using the combination probability but then why is this method using the binomial wrong??
This was a tricky question.View attachment 59314
w12-63
Please explain how to do this question :/
The ms is toooo vague saying (2/3)^7
Why and how??
Nope.Now take the cross product of both to get the normal of 'q'. (you know how to do that right?)
This was a tricky question.
Imagine that the first tile placed could be any tile from the three colours mentioned. The probability of it will be 1/3. Then, the probability that the next tile is difference from the previous one is 2/3. The probability that the 3rd tile will be different from the 2nd tile will also be 2/3. Continue this upto 8 tiles. As these events are independent, multiply all the probabilities. Since the first tile can be any tile from the three, we'll multiply with 3. You'll get:
3 * 1/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 = (2/3)^7 Ans.
Ooooh thanks a lot!This was a tricky question.
Imagine that the first tile placed could be any tile from the three colours mentioned. The probability of it will be 1/3. Then, the probability that the next tile is difference from the previous one is 2/3. The probability that the 3rd tile will be different from the 2nd tile will also be 2/3. Continue this upto 8 tiles. As these events are independent, multiply all the probabilities. Since the first tile can be any tile from the three, we'll multiply with 3. You'll get:
3 * 1/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 = (2/3)^7 Ans.
Nope.
Thanks for clearing out the other doubts.
ThanksHere you go
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Never mind the pencil mapping ... it's messy.
No problem.
This was the same type of question you had asked me to do binomial expansion in remember?
Please help asap, I have test in my tuition tomorrow early morning. I rememberThis was the same type of question you had asked me to do binomial expansion in remember?
Well, I can't believe I'm getting it wrong again! So frustrating.
nehaoscar please once again?![]()
Got it bro!This was the same type of question you had asked me to do binomial expansion in remember?
Well, I can't believe I'm getting it wrong again! So frustrating.
nehaoscar please once again?![]()
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