sandipgumtya wrote:Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15?
A. 5/16
B. 15/32
C. 1/2
D. 21/32
E. 11/16
In the beginning, Kate has $15.
For every tails, Kate gets $1; for every heads, she loses $1.
Implication:
To finish with less than $15, Kate must lose AT LEAST ONCE, implying that AT LEAST ONE of the 5 flips must be heads.
Test outcomes with at least one heads:
4 tails, 1 heads --> Kate has 10+4-1 = $13
3 tails, 2 heads --> Kate has 10+3-2 = $11
2 tails, 3 heads --> Kate has 10+2-3 = $9.
Only the options in red yield for Kate more than $10 but less than $15.
Question stem, rephrased:
If a coin is flipped 5 times, what is the probability that the coin lands on tails exactly 4 times or exactly 3 times?
P(exactly n times) = P(one way) * total possible ways.
Case 1: Exactly 4 tails
P(one way):
in 5 tosses, one way to get exactly 4 tails is TTTTH.
P(TTTTH) = (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32.
Total possible ways:
Any arrangement of the letters TTTTH represents one way to get exactly 4 T"s and 1 H.
Thus, to account for ALL OF THE WAYS to get exactly 4 T's and 1 H, the result above must be multiplied by the number of ways to arrange the letters TTTTH.
Number of ways to arrange 5 elements = 5!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 4! to account for the four identical T's:
5!/4! = 5.
Multiplying the results above, we get:
P(exactly 4 T's) = 5 * 1/32 = 5/32.
Case 1: Exactly 3 tails
P(one way):
in 5 tosses, one way to get exactly 3 tails is TTTHH.
P(TTTHH) = (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32.
Total possible ways:
Any arrangement of the letters TTTHH represents one way to get exactly 3 T"s and 2 H's.
Thus, to account for ALL OF THE WAYS to get exactly 3 T's and 2 H's, the result above must be multiplied by the number of ways to arrange the letters TTTHH.
Number of ways to arrange 5 elements = 5!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 3! to account for the three identical T's and by 2! to account for the two identical H's:
5!/(3!2!) = 10.
Multiplying the results above, we get:
P(exactly 3 T's) = 10 * 1/32 = 10/32.
Thus:
P(exactly 4 tails or exactly 3 tails) = 5/32 + 10/32 = 15/32.
The correct answer is
B.
More practice:
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https://www.beatthegmat.com/rain-check-t79099.html
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