Suppose a coin has PH 23 and PH 13 a For five tosses of th

Suppose a coin has P(H) = 2/3 and P(H) = 1/3.

a)   For five tosses of this coin find the probability of getting exactly 4 heads and 1 tail.

b)    For five tosses of this coin find the probability of getting exactly 2 tails and 3 heads.

Solution

Do you mean P(H) = 2/3, and P(Tails) = 1/3?

Here, we treat \"success\" as getting heads.

If so, this is the solution:

a)

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    5      
p = the probability of a success = 2/3 =   0.666666667      
x = the number of successes =    4      
          
Thus, the probability is          
          
P (    4   ) =    0.329218107 [ANSWER]

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b)

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    5      
p = the probability of a success =    0.666666667      
x = the number of successes =    3      
          
Thus, the probability is          
          
P (    3   ) =    0.329218107 [ANSWER]