Problems top top coin toss probability are described here with different examples.

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When us flip a coin over there is always a probability to get a head or a tail is 50 percent.Suppose a coin tossed climate we obtain two feasible outcomes one of two people a ‘head’ (H) or a ‘tail’ (T), and it is impossible to predict whether the result of a toss will be a ‘head’ or ‘tail’.

The probability for equally likely outcomes in an event is:

Number that favourable outcomes ÷ Total number of possible outcomes


Total variety of possible outcomes = 2

(i) If the favourable result is head (H).

Number that favourable outcomes = 1.

Therefore, P(getting a head)

number of favorable outcomes = P(H) = total number of possible outcomes

= 1/2.

(ii) If the favourable result is tail (T).

Number the favourable outcomes = 1.

Therefore, P(getting a tail)

variety of favorable outcomes = P(T) = total variety of possible outcomes

= 1/2.

Word difficulties on Coin Toss Probability:

1. A coin is tossed double at random. What is the probability the getting

(i) at the very least one head

(ii) the very same face?

Solution:

The feasible outcomes space HH, HT, TH, TT.

So, total variety of outcomes = 4.

(i) number of favourable outcomes for event E

                              = number of outcomes having actually at least one head

                              = 3 (as HH, HT, TH are having actually at least one head).

So, by definition, P(F) = \(\frac34\).

(ii) variety of favourable outcomes for event E

                              = variety of outcomes having actually the very same face

                              = 2 (as HH, TT are have the very same face).

So, by definition, P(F) = \(\frac24\) = \(\frac12\).

2.

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 If three fair coins are tossed randomly 175 times and also it is discovered that three heads appeared 21 times, two heads appeared 56 times, one head appeared 63 times and also zero head showed up 35 times. 

What is the probability the getting 

(i) 3 heads, (ii) 2 heads, (iii) one head, (iv) 0 head. 

Solution: 

Total variety of trials = 175. 

Number the times 3 heads showed up = 21. 

Number the times 2 heads showed up = 56. 

Number of times one head appeared = 63. 

Number of times zero head showed up = 35. 

Let E1, E2, E3 and E4 it is in the events of gaining three heads, 2 heads, one head and also zero head respectively.

(i)P(getting three heads)

variety of times three heads appeared = P(E1) = total variety of trials

= 21/175

= 0.12

(ii) P(getting two heads)

variety of times 2 heads showed up = P(E2) = total number of trials

= 56/175

=0.32

(iii) P(getting one head)

number of times one head appeared = P(E3) = total number of trials

=63/175

= 0.36

(iv) P(getting zero head)

number of times zero head showed up = P(E4) = total number of trials

= 35/175

= 0.20

Note: Remember when 3 coinsare tossed randomly, the only feasible outcomes

space E2, E3, E4 andP(E1) + P(E2) + P(E3) + P(E4)

= (0.12 + 0.32 + 0.36 + 0.20)

= 1

3. 2 coins space tossed randomly 120 times and also it is uncovered that 2 tailsappeared 60 times, one tailappeared 48 times and no tail appeared 12 times.

If two coins room tossed in ~ random, what is theprobability of obtaining

(i) 2 tails,

(ii) 1 tail,

(iii) 0 tail

Solution:

Total number oftrials = 120

Number of times 2 tails appear= 60 

Number of times 1 tail appears= 48

Number of time 0 tail appears= 12

Let E1, E2 and E3 it is in the occasions of gaining 2 tails, 1 tail and 0 tail respectively.

(i) P(getting2 tails)

number of times 2 tails appear = P(E1) = total variety of trials

= 60/120

= 0.50

(ii) P(getting 1 tail)

number of times 1 tail appear = P(E2) = total variety of trials

= 48/120

= 0.40

(iii) P(getting0 tail)

number of times no tail appear = P(E3) = total number of trials

= 12/120

= 0.10

Note:

Remember while tossing 2 coins simultaneously, the only feasible outcomes are E1, E2, E3 and, P(E1) + P(E2) + P(E3)

= (0.50 + 0.40 + 0.10)

= 1


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4. Suppose a fair coin is randomlytossed for 75 times and also it is uncovered that head turns up 45times and tail 30 times. What is the probability of acquiring (i)a head and also (ii) a tail?

Solution:

Total number of trials = 75.

Number of times head transforms up = 45

Number of time tail turns up = 30

(i) let X it is in the occasion ofgetting a head.

P(getting a head)

number of times head turns up = P(X) = total variety of trials

= 45/75

= 0.60

(ii) let Y bethe event of getting a tail.

P(getting a tail)

number of times tail transforms up = P(Y) = total variety of trials

= 30/75

= 0.40

Note: Remember as soon as afair coin is tossed and also then X and Y arethe only feasible outcomes, and also