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An unbiased coin is tossed 5 times. Suppose that a variable X is assigned the value k when k consecutive heads are obtained for k = 3, 4, 5, otherwise X takes the value -1. Then, the expected value of X, is:
a). $ \dfrac{1}{8} $
b). $ \dfrac{3}{16} $
c). $ -\dfrac{1}{8} $
d). $ -\dfrac{3}{16} $

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Last updated date: 18th Jun 2024
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Answer
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Hint:
 We know from the question that the coin is tossed 5 times. So, we can get 0, 1, 2, 3, 4, 5 different types of consecutive held. So, we will first find the probability of each type of consecutive head and then we will calculate the value of X by multiplying the value it takes with the respective probability occurrence and then summing all of them.

Complete step by step answer:
We know from the question that the coin is tossed 5 times. And, we also know that when we toss a fair coin then we two possible outcomes, that is head, and tail.
So, total number of possible combinations we will get when coin is tossed 5 times is $ 2\times 2\times 2\times 2\times 2 $ = 32
Also, we will have chances of getting 0, 1, 2, 3, 4, or 5 consecutive heads when coin is tossed 5 times.
Now, we will calculate the probability of each of them by using grouping.
So, probability of getting ‘0’ consecutive head (i.e. TTTTT) is $ \dfrac{1}{32} $ .
Probability of getting ‘1’ consecutive head (i.e. HTTTT, THTTT, TTHTT, TTTHT, TTTTH, HTHTT, HTTHT, HTTTH, THTHT, TTHTH, THTTH ) is $ \dfrac{12}{32} $
Probability of getting ‘2’ consecutive head (i.e. HHTTT, THHTT, TTHHT, TTTHH, HHTHH, HHTTH, HHTHT, THTHH, HTTHH, THHTH, HTHHT ) is $ \dfrac{11}{32} $ .
Probability of getting ‘3’ consecutive head (i.e. HHHTT, THHHT, TTHHH, HTHHH, HHHTH ) is $ \dfrac{5}{32} $
Probability of getting ‘4’ consecutive head (i.e. HHHHT, THHHH ) is $ \dfrac{2}{32} $ .
Probability of getting ‘5’ consecutive head (i.e. HHHHH ) is $ \dfrac{1}{32} $ .
Now, we know from the question that X is assigned value k when k consecutive heads are obtained for k = 3, 4, 5, and X takes value -1 when k = 0, 1, or 2.
So, we can say that when k = 3, X will take value $ 3\times \Pr \text{obability of getting 3 consecutive heads} $ = $ 3\times \dfrac{5}{32} $ and similarly we can do for 4 and 5.
And, when k = 0, X will take value $ -1\times \Pr \text{obability of getting 1 consecutive head} $ = $ -1\times \dfrac{1}{32} $ and similarly we can do for 1 and 2
So, total value of X will be = $ -1\times \dfrac{1}{32}-1\times \dfrac{12}{32}-1\times \dfrac{11}{32}+3\times \dfrac{5}{32}+4\times \dfrac{2}{32}+5\times \dfrac{1}{32} $
 $ =-\dfrac{1}{32}-\dfrac{12}{32}-\dfrac{11}{32}+\dfrac{15}{32}+\dfrac{8}{32}+\dfrac{5}{32} $
 $ =\dfrac{28-24}{32} $
 $ =\dfrac{4}{32}=\dfrac{1}{8} $
This is our required solution.
So, option (a) is our correct answer.

Note:
Which calculating the value of X students are required to note that there is always the probability of obtaining that value for example let us say that X is assigned the value 3, so there is always the probability of getting 3 consecutive head which is less than 1 so we are required to multiply probability to an assigned value which is here 3 to calculate the value of X.


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