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Difficulty:
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Question Stats:
66% (02:05) correct
34%
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wrong
based on 268
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An unfair coin comes up heads 60% of the time and tails 40% of the time when it is tossed. What is the probability of getting exactly three heads on five tosses of this coin?
A. \(\frac{10}{32}\)
B. \(\frac{27}{125}\)
C. \(\frac{2^33^3}{5^5}\)
D. \(\frac{3}{5}\)
E. \(\frac{6^3}{5^4}\)
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A. \(\frac{10}{32}\)
B. \(\frac{27}{125}\)
C. \(\frac{2^33^3}{5^5}\)
D. \(\frac{3}{5}\)
E. \(\frac{6^3}{5^4}\)
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HHHTT
6^3/10^3 * 4^3/10^3 * 5!/3!*2!
= 6^3/5^4.
6^3/10^3 * 4^3/10^3 * 5!/3!*2!
= 6^3/5^4.
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total fair ways ; 5!/3!*2! ; 10
and for heads and tails ; 5c3*(.6)^3 *(.4)^2*5c2
total P of exactly three heads on five tosses ; 5c3*(.6)^3 *(.4)^2*5c2/10
solve =
\(\frac{6^3}{5^4}\)
IMO E
and for heads and tails ; 5c3*(.6)^3 *(.4)^2*5c2
total P of exactly three heads on five tosses ; 5c3*(.6)^3 *(.4)^2*5c2/10
solve =
\(\frac{6^3}{5^4}\)
IMO E
Bunuel
60℅=3/5
So, 3 times heads and 2 times heads => (3/5)*(3/5)*(3/5)*(2/5)*(2/5)
Now Here order matters because first throw can result in tail and second throw can result in head and so on......
So We will multiply by 5!/(3!*2!)
Thus answer E.
Posted from my mobile device
So, 3 times heads and 2 times heads => (3/5)*(3/5)*(3/5)*(2/5)*(2/5)
Now Here order matters because first throw can result in tail and second throw can result in head and so on......
So We will multiply by 5!/(3!*2!)
Thus answer E.
Posted from my mobile device
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