- \(\dfrac{15}{128}\)
- \(\dfrac{15}{32}\)
- \(\dfrac{15}{64}\)
- \(\dfrac{30}{32}\)
Answer (Detailed Solution Below)
Option 3 : \(\dfrac{15}{64}\)
Free
10 Questions 10 Marks 6 Mins
Calculation:
The total number of possible outcome = 26 = 64
The total possibilities of getting only 4 head = 15
The required probability = 15 / 64
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When in doubt about an exercise, a reasonable mathematical approach is to simplify the exercise data if possible; here, let's look for the probability of getting exactly $2$ heads when a coin is tossed $a=3$ times. Let's describe this random experience with the sample space $S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$.
By the way, we make sure that $Card(S)=2^a=2^3=8$. The event $A':$"getting exactly $2$ heads" is $A'=\{HHT,HTH,THH\}$. We have$Card(A')=\mathrm{C}_{3}^{2}=\mathrm{C}_{3}^{1}=3$. Suppose that the outcomes in the sample space are equally likely to occur. Then $p(A')=\frac{\mathrm{C}_{3}^{2}}{2^3}=\frac{\mathrm{C}_{3}^{1}}{2^3}$.
So back to our original exercise, and based on what we have learned in our simple case, $p(A)=\frac{\mathrm{C}_{6}^{4}}{2^6}=\frac{\mathrm{C}_{6}^{2}}{2^6}=\frac{\frac{6\times5}{2!}}{2^6}$.
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Updated On: 27-06-2022
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