A fair coin is tossed 5 times. What is the probability that it lands

5/16

Two heads can be in a toss of 5 times in 5C2 = 10 ways

Each way(number of times at least two heads will come) has a probability of 1/32.

Hence 10*1/32 = 5/16

It is B....

Total possible outcomes=2*2*2*2*2=32
Favourable outcomes=5C2=10

Probability=10/32=5/16....

Yes.......
Calculated ofr only two heads.....

Favorable outcomes are 5C2+5C3 + 5C4+5C5
=> 10+10+5+1
=> 26
26/32 = 13/16

Hence D

I don't understand whu you sum up "10+10+5+1".

Can someone explain it?

We sum up the number of ways you can toss heads up 2 times, 3 times, 4 times and 5 times....

Probability of all tails = 1/32
Probability of 4 tails = 5/32
------------------------------------------------------------------------------
Probability of all tails or 4 tails = 1/32 + 5/32 = 6/32 = (3/16)
------------------------------------------------------------------------------

Therefore probability of at least 2 heads = 1- Prob of all or 4 tails = 1- (3/16) = 13/16

Hence (D)

Fair coin is tossed 5 times . Hence total number of outcomes = 2^5 = 32.

Problem asks for probability of getting atleast heads twice. Hence if we calculate probability of getting Heads exactly once and probability of not getting Heads at all and subract it from the total probability of the event which is 1 (As total probability of certain event will be always 1) we can get the probability of getting atleast heads twice.

Probability of getting exactly one head and no heads= (Number of possible outcomes [ HTTTT , THTTT, TTHTT, TTTHT, TTTTH , TTTTT] = 6)/ (Total possible outcomes = 32)

=> 6/32 = 3/16

Hence probability of getting atleast heads twice = 1 - (3/16) = 13/16 => Choice [D]

Sol:

2H3T, 3H2T, 4H1T, 5H calculating this will take longer than calculating 5T, 4T1H

lets calculate 5T, 4T1H

5T= 1/2^5 = 1/32
4T1H = (1/32) (5!/4!) = 5/32

5T+4T1H= 6/32 = 3/16

1- (5T+4T1H) = 1-3/16 = 13/16

(D)

Here are the numbers to remember:

1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

The first row applies to 3 flip questions, the second to 4 flip questions and the third to 5 flip questions.
If we want at least 2 heads, we want 2 heads, 3 heads, 4 heads OR 5 heads. Pretty much whenever we see "OR" in probability, we add the individual probabilities.

Looking at the 5 flip row, we have 1 5 10 10 5 1. For 2H, 3H, 4H and 5H we add up the 3rd, 4th, 5th and 6th entries: 10+10+5+1=26.

Summing the whole row, we get 32.

So, the chance of getting at least 2 heads out of 5 flips is 26/32 = 13/16.

Here are the numbers to remember:

1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

The first row applies to 3 flip questions, the second to 4 flip questions and the third to 5 flip questions.
If we want at least 2 heads, we want 2 heads, 3 heads, 4 heads OR 5 heads. Pretty much whenever we see "OR" in probability, we add the individual probabilities.

Looking at the 5 flip row, we have 1 5 10 10 5 1. For 2H, 3H, 4H and 5H we add up the 3rd, 4th, 5th and 6th entries: 10+10+5+1=26.

Summing the whole row, we get 32.

So, the chance of getting at least 2 heads out of 5 flips is 26/32 = 13/16.

For all these kind of questions we can answer using this concept ( hope all of you might have studied during your school days):

Probability of 'r' success out of 'n' trials = = nCr (P)^r (Q)^n-r

P - probability of success
Q - probability of failure

According to this question :
We need to find probability of head >= 2 out of 5 trials.
P - probability of success = 1/2(probability of getting head out of a toss)
Q - probability of failure= 1/2(probability of getting tail out of a toss)

In mathematical form:
P(H>=2) = 1-P(H<2) = 1- (P( H=0) + P( H=1)).

P(H =0) = 5C0 (1/2)^0 (1/2)^5 = (1/2)^5---------- eqn 1

P(H =1) = 5C1(1/2)^1 (1/2)^4 = (5/2)^5---------- eqn 2

Finally,

1-P(H<2) = 1- (P( H=0) + P( H=1)) = 1 - ((1/2)^5)+(5/2)^5) = 1- (6 / 2^5) = 26/32= 13/16.


Final note, though this method looks lengthy, it will be applicable to all atleast,atmost and exactly equal to 'n' of questions in probability.

In GMAT, more than solving fast we have to keep our mind unstressed and fresh. So that we can be active for such a long exam.

Hope it helps [WINKING FACE]

Fair coin is tossed 5 times.total number of outcomes = 2^5 = 32

5C2 + 5C3 +5C4 + 5C5=10+10+5+1=26

26/32=13/16

Ans-D

prob = 1 -(heads at max once)
= 1 - (heads exactly zero times or heads exactly one times)

at max once means two cases:
case 1 : exactly zero heads
=> TTTTT
=> 1/32

case 2 : heads exactly once
=> HTTTT
=> (5!/4!)/32
=> 5/32

Now after putting the values
=> 1 -(heads at max once)
=> 1 - ((1/32) + (5/32))
=> 13/16

We will have the following cases: (2H, 3T); (3H, 2T); (4H, 1T); (5H)

Also, HHHTT is different from HTTHT. Therefore, for (2H, 3T) => (1/2)^5 * 5!/(3! * 2!) = 10/32

(3H, 2T) => 10/32 ; (4H, 1T) => 5/32 ; (5H) => 1/32

Total = 13/16. Option D.

It is more easy way to with complement (at least 2 heads):

total - one head + no head
total = 2^5 = 32

one head = 5C1 + 5C0 = 5 + 1 = 6

Use complement here = 32 - 6 = 26/32 = 13/16.
I hope you understand this is most easiest way to do

Total possible events = 2 x 2 x 2 x 2 x 2 = 32
Likely events = Heads up at least twice.
Heads up at least twice = Total events - No heads two-time
No heads two-time = 6 [HTTTT, THTTT, TTHTT, TTTHT, TTTTH, TTTTT]
Heads up at least twice = 32 - 6 = 26
The probability that it lands heads up at least twice = 26/32 = 13/16
The final answer is d.

Given that A fair coin is tossed 5 times and we need to find What is the probability that it lands heads up at least twice

P(At least twice heads) = P(2H) + P(3H) + P(4H) + P(5H) = 1 - P(0H) - P(1H)

P(0H)

Total number of cases = \(2^5\) = 32
Case in which we get 0H or 5T is TTTTT => 1

=> P(0H) = \(\frac{1}{32}\)

P(1H)

Case in which we get 1H can be found by putting H in any of the 5 slots _ _ _ _ _ => HTTTT, THTTT, TTHTT, TTTHT, TTTTH => 5

=> P(1H) = \(\frac{5}{32}\)

P(At least twice heads) = 1 - P(0H) - P(1H) = 1 - \(\frac{1}{32}\) - \(\frac{5}{32}\) = \(\frac{32 - 6}{32}\) = \(\frac{26}{32}\) = \(\frac{13}{16}\)

So, Answer will be D
Hope it helps!

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