Probability of having at least 2 heads = 1 - (probability of 1 heads + prob of no heads)

1 - {5C1(1/2)^1 (1/2)^4 + (1/2)^5}

1- {5/32 + 1/32} = 26/32

= 13/16

Answer D

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P of no head = 1/32
P of 1 head = 5c1* 1/16*1/2 ; 5/32
P of atmost 1 head = 1/32+ 5/32 ; 6/32 ; 3/16
atleast 2 times ; 1-3/16 ; 13/16
IMO D

What is the probability of tossing a coin five times and having heads appear at least two times?

(A) 1/32
(B) 1/16
(C) 15/16
(D) 13/16
(E) 7/8

For a head to appear at least two times, we have 5C2 + 5C3 + 5C4 + 5C5 = 10+10+5+1=26 total possibilities
Total events = 2^5 = 32
P(at least 2H) = 26/32 = 13/16.

The answer is therefore D.

A coin is tossed 5 times,
possibility each time: 2 (1 head, 1 tail)
=> total outcomes = 2^5 = 32.

minimum 2 heads means: (1) HHTTT, (2) HHHTT, (3) HHHHT and (4) HHHHH

1. total possibilities of HHTTT: 5! / (2! x 3!) = 10 nos
2. total possibilities of HHHTT: 5! / (3! x 2!) = 10 nos
3. total possibilities of HHHHT: 5! / (4! x 1!) = 5 nos
4. total possibilities of HHHHH: 5! / (5! x 0!) = 1 nos

=> total possibilities of minimum 2 heads = 10+10+5+1 = 26 nos

=> P ( min. 2 heads) = favourable outcomes / total outcomes
= 26/32
= 13/16

Answer is (D).

Asked: What is the probability of tossing a coin five times and having heads appear at least two times?

Total cases = 2^5 = 32
Unfavorable cases = Head not appearing + Head appearing once = 1 + 5 = 6
Favorable cases = 32 - 6 = 26

Probability = Favorable cases / Total cases = 26/32 = 13/16

IMO D

We need to determine the probability of seeing heads 2, 3, 4, or 5 times when a coin is flipped 5 times.

It is easier to calculate the opposite event and subtract it from 1.

Opposite event = probability of seeing 0 head or probability of seeing 1 head

Total number of outcomes = 2x2x2x2x2 = 32

Favourable outcomes :
0 Head i.e TTTTT =1 event
1 Head i.e HTTTT = 5 events since H can be arranged anywhere

Probability of seeing 0 head or probability of seeing 1 head \(= \frac{1}{32} + \frac{5}{32}\)

\(P = 1 - \frac{6}{32} =\frac{13}{16}\)