When a certain coin is flipped, the probability that the coin will lan

Question

When a certain coin is flipped, the probability that the coin will land on head or tail is 1/2 each. If the coin is flipped 4 times, what is the probability that it will land on tail at least twice on 4 flips?

A. \(\frac{3}{8}\)
B. \(\frac{1}{16}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{8}\)
E. \(\frac{11}{16}\)

A.  \frac{3}{8}  
B.  \frac{1}{16}  
C.  \frac{1}{2}  
D.  \frac{5}{8}  
E.  \frac{11}{16}

attari92 · Answer

total no. of outcomes =16
favorable outcomes =(2h2t) 4!/2!*2! =6; (3t1h) 4!/3!=4; (4t) 4!/4!=1. Total favorable outcomes =11
req. probability=11/16-> E

Hatakekakashi · Answer

It must land on tail atleast twice = 1- land of tail at most once

HHHH = 1
THHH = 4 ways this can happen
total ways = 1+4

Number of outcomes = 2^4 =16
P(at most once) = 5/16

P( atleast twice ) = 1 - P( at most once) =1 - 5/16 = 11/16

MathRevolution · Answer

==> In general, you solve probability questions using nCr combination. In other words, from TTHH, there are (4!/2!/2!)=6 possibilities, so you get (1/2)(1/2)(1/2)(1/2)*6=3/8. From TTTH, there are (4!/3!)=4 possibilities, so you get (1/2)(1/2)(1/2)(1/2)*4=1/4. From TTTT, there is only 1 possibility, so you get (1/2)(1/2)(1/2)(1/2)=1/16 and (3/8)+(1/4)+(1/16)=(6+4+1)/16=11/16.

The answer is E.
Answer: E

The answer is E. 
Answer: E

Gijmaja · Answer

probability that no tail= (1/2)^4
probability that 1 tail= 4C1*(1/2)^4

1-(1/2)^4-4C1*(1/2)^4=11/16
Answer E.

SaquibHGMATWhiz · Answer

Solution:

Two ways to go about this:  Calculating the favorable cases and adding them up  or  Calculating the unfavorable cases and subtracting them from 1  
 Let us look at the  latter method   
 So, we need ATLEAST 2 tails. So, let's talk about getting 1 tail and no tail  
 1 tail:
  We need to rearrange THHH which can be done in  \frac{4!}{3!}=4  ways   Probability of each tail or head is  \frac{1}{2}    So, probability of getting 1 tail  =4	imes (\frac{1}{2})^4=\frac{4}{16}=\frac{1}{4}    
 2 tails:
  We need to rearrange HHHH which can be done in  \frac{4!}{4!}=1  ways   Probability of each tail or head is  \frac{1}{2}    So, probability of getting 1 tail  =1	imes (\frac{1}{2})^4=\frac{1}{16}   
 
 Probability of at least 2 tails  =1-(\frac{1}{4}+\frac{1}{16})=1-\frac{5}{16}=\frac{11}{16}

Hence the right answer is    Option E

BrushMyQuant · Answer

When a certain coin is flipped, the probability that the coin will land on head or tail is 1/2 each. If the coin is flipped 4 times

Coin is tossed 4 times => Total number of cases =  2^4  = 16

What is the probability that it will land on tail at least twice on 4 flips?

=> Number of cases for at least 2 Tail = 16 - (0 Tail + 1 Tail) = 16 - (1 + 4)  
=> Number of cases for at least 2 Tail = 16 - 5 = 11

P(At least 2 T) =  \frac{11}{16}

So,  Answer will be E 
Hope it helps!

Playlist on Solved Problems on Probability  here

Watch the following video to MASTER Probability with Coin Toss Problems

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

messi16 · Answer

Every flip has 2 possible choices, and there are four flips. Thus, we know the number of total outcomes is 2^4 = 16. Then, for the various ways we could have tails, we have 4 C 2, 4 C 3, and 4 C 4, which is simply 6 + 4 + 1 = 11. 11/16 is our final answer.

When a certain coin is flipped, the probability that the coin will lan

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When a certain coin is flipped, the probability that the coin will lan

Prep Toolkit

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