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A coin is weighted so that the probability of heads on any flip is 0.6, while the probability of tails is 0.4. If the coin is flipped 5 times independently, which of the following represents the probability that tails will appear no more than twice?
A. \((0.6)^5 + 5(0.6)^4(0.4) + 10(0.6)^3(0.4)^2\)
B. \((0.6)^5 + 4(0.6)^4(0.4) + 6(0.6)^3(0.4)^2\)
C. \((0.6)^5 + 3(0.6)^4(0.4) + 2(0.6)^3(0.4)^2\)
D. \((0.6)^5 + 2(0.6)^4(0.4) + (0.6)^3(0.4)^2\)
E. \((0.6)^5 + (0.6)^4(0.4) + (0.6)^3(0.4)^2\)
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A. \((0.6)^5 + 5(0.6)^4(0.4) + 10(0.6)^3(0.4)^2\)
B. \((0.6)^5 + 4(0.6)^4(0.4) + 6(0.6)^3(0.4)^2\)
C. \((0.6)^5 + 3(0.6)^4(0.4) + 2(0.6)^3(0.4)^2\)
D. \((0.6)^5 + 2(0.6)^4(0.4) + (0.6)^3(0.4)^2\)
E. \((0.6)^5 + (0.6)^4(0.4) + (0.6)^3(0.4)^2\)
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Bunuel
MANHATTAN GMAT OFFICIAL SOLUTION:
In this probability problem, we first need to split up the successful cases into three batches:
a) Flipping zero tails
b) Flipping exactly one tails
c) Flipping exactly two tails
a) Flipping zero tails is the same as flipping five heads in a row (HHHHH). Since the probability of flipping one heads is 0.6, and the flips are independent, we get \((0.6)^5\). Unfortunately, this term is common to every answer choice, perhaps not surprisingly.
b) Flipping exactly one tails is the same as flipping four heads and one tails. These can come in any of five orders, depending on the position of the “odd man out”:
THHHH
HTHHH
HHTHH
HHHTH
HHHHT
Each of these sub-cases has a probability of \((0.6)^4*(0.4)\). Since we have five separate sub-cases, any of which counts as success, we add up the probabilities to get \(5*(0.6)^4*(0.4)\). There is only one answer choice that fits: we could stop here.
For the sake of completeness, let’s do the last part.
c) Flipping exactly two tails is the same as flipping three heads and two tails in any order. To count the arrangements, we can either write them all out or use a combinatorics approach, such as the anagram method. Using the latter, we get \(\frac{5!}{(3!2!)}= 10\) arrangements. This 10 becomes the coefficient on the probability of one particular arrangement (such as HHTHT). We get \(10*(0.6)^3*(0.4)^2\) as the last term.
The correct answer is A.
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Bunuel
Probability of Head, P(H) = 0.6
Probability of Tail, P(T) = 0.4
Tail will appear NO more than twice
i.e. favourable cases
2 Tails and 3 Heads, Probability = 5C2*(0.6)^3*(0.4)^2
1 Tail and 4 Heads, Probability = 5C1*(0.6)^4*(0.4)^1
0 Tail and 5 Heads, Probability = (0.6)^5
Required Probability = Sum of all Favourable cases = (0.6)^5 + 5(0.6)^4(0.4) + 10(0.6)^3(0.4)^2
Answer: option A
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General Discussion
Posts: 2
Required prob -
THHHH - 5!/(4!) (0.4)(0.6)^4
TTHHH - 5!/2!.3! (0.4)^2(0.6)^3
HHHHH-5!/5! (0.6)^5
The sum matches (A)
THHHH - 5!/(4!) (0.4)(0.6)^4
TTHHH - 5!/2!.3! (0.4)^2(0.6)^3
HHHHH-5!/5! (0.6)^5
The sum matches (A)
Answer is A
3 cases possible:
hhhhh or hhhht (5 cases) or hhhtt(10 cases)
Multiplying the respective probabilities with the weights gives us A
3 cases possible:
hhhhh or hhhht (5 cases) or hhhtt(10 cases)
Multiplying the respective probabilities with the weights gives us A
GMATinsight
Hi,
i think it is a typing error. pl edit it..
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Bunuel
the formula for these Qs is..
nCr*(p)^r*(1-p)^(n-r)...
A is the answer
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Posts: 1,348
Bunuel
ALWAYS LOOK AT THE ANSWER CHOICES.
P(no more than two tails) = P(all heads) + P(4 heads and 1 tails) + P(3 heads and 2 tails)
P(heads) = 0.6
P(tails) = 0.4
Thus, the second term in each answer choice must represent P(4 heads and 1 tails):
(0.6)⁴(0.4)
Since the 1 tails can happen on any of the 5 flips -- implying 5 different ways to get exactly 1 tails -- the second term must include a factor of 5:
5(0.6)⁴(0.4)
Only A satisfies this condition.
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Given that A coin is weighted so that the probability of heads on any flip is 0.6, while the probability of tails is 0.4. If the coin is flipped 5 times independently. We need to find which of the following represents the probability that tails will appear no more than twice?
As the coin is tossed five times then number of cases = \(2^5\) = 32
P(Tails will appear no more than twice) = P(0T) + P(1T) + P(2T)
P(0T)
P(0T) = P(HHHHH) = 0.6*0.6*0.6*0.6*0.6 = \((0.6)^5\)
P(1T)
Now out of the five place _ _ _ _ _ Tail can come in any of the places in 5C1 ways = 5 ways
=> P(1T) = Number of places * P(Tail) * P(Head) * P(Head) * P(Head) * P(Head) = 5*0.4*0.6*0.6*0.6*0.6 = \(5(0.6)^4(0.4)\)
P(2T)
Now out of the five place _ _ _ _ _ we need to find two places where Tail can come => 5C2 ways = \(\frac{5!}{2!*(5-2)!}\) ways = \(\frac{5*4*3!}{2!*3!}\) ways = 10 ways
=> P(2T) = Number of places * P(Tail) * P(Tail) * P(Head) * P(Head) * P(Head) = 10*0.4*0.4*0.6*0.6*0.6 = \(10(0.6)^3(0.4)^2\)
P(Tails will appear no more than twice) = P(0T) + P(1T) + P(2T) = \((0.6)^5\) + \(5(0.6)^4(0.4)\) + \(10(0.6)^3(0.4)^2\)
So, Answer will be A
Hope it helps!
Watch the following video to learn How to Solve Probability with Coin Toss Problems
_________________
As the coin is tossed five times then number of cases = \(2^5\) = 32
P(Tails will appear no more than twice) = P(0T) + P(1T) + P(2T)
P(0T)
P(0T) = P(HHHHH) = 0.6*0.6*0.6*0.6*0.6 = \((0.6)^5\)
P(1T)
Now out of the five place _ _ _ _ _ Tail can come in any of the places in 5C1 ways = 5 ways
=> P(1T) = Number of places * P(Tail) * P(Head) * P(Head) * P(Head) * P(Head) = 5*0.4*0.6*0.6*0.6*0.6 = \(5(0.6)^4(0.4)\)
P(2T)
Now out of the five place _ _ _ _ _ we need to find two places where Tail can come => 5C2 ways = \(\frac{5!}{2!*(5-2)!}\) ways = \(\frac{5*4*3!}{2!*3!}\) ways = 10 ways
=> P(2T) = Number of places * P(Tail) * P(Tail) * P(Head) * P(Head) * P(Head) = 10*0.4*0.4*0.6*0.6*0.6 = \(10(0.6)^3(0.4)^2\)
P(Tails will appear no more than twice) = P(0T) + P(1T) + P(2T) = \((0.6)^5\) + \(5(0.6)^4(0.4)\) + \(10(0.6)^3(0.4)^2\)
So, Answer will be A
Hope it helps!
Watch the following video to learn How to Solve Probability with Coin Toss Problems
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