Kevin flips a coin four times. WHat is the probability the he gets hea

Question

Kevin flips a coin four times. What is the probability the he gets heads on at least one of the four flips?

A. 1/16 
B. 1/4 
C. 3/4 
D. 13/16 
E. 15/16

eswarchethu135 · Answer

P(heads on at least one of the four flips) = 1- P(heads on none of the four flips)

=  1 - (\frac{1}{2})^4

=  1 - \frac{1}{16}

=  \frac{15}{16}

OPTION:  E

LevanKhukhunashvili · Answer

Easier to find the prob for not getting head at all and then find the prob of getting at least one head by 1-(prob of getting no head)

Prob of getting only tails:  (\frac{1}{2})^4 
prob of getting at least one head 1- \frac{1}{16} = \frac{15}{16} 
IMO
Ans:  E

saukrit · Answer

Getting Head on at least one flip of four=  P(1 Head) + P(2 Head) + P(3 Head) + P(4 Head)

Also,  P (0 Head) + P(1 Head) + P(2 Head) + P(3 Head) + P(4 Head)  = 1

Therefore, Getting Head on at least one flip of four=  P(1 Head) + P(2 Head) + P(3 Head) + P(4 Head) = 1- P (0 Head)

P (0 Head) = (1/2)*(1/2)*(1/2)*(1/2) = (1/16)

Probability of getting Head on at least one flip of four  =1- P (0 Head) = 1- (1/16) =   (15/16)

IMO, Answer is E

BrushMyQuant · Answer

P(0H) + P(1H) + P(2H) + P(3H) + P(4H) = 1 
=> P(1H) + P(2H) + P(3H) + P(4H) = 1 - P(0H)

P(At least 1 H) = 1 - P(0H)

1 - P(4T) = 1 -  \frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}  = 1 -  \frac{1}{16}  =  \frac{15}{16}

So,  Answer will be E 
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

https://www.youtube.com/watch?v=LRURv38kZw8

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