Three coins are tossed simultaneously $500$ times with the following frequencies of different outcomes:
Number of heads $0$ $1$ $2$ $3$ Frequency $70$ $190$ $175$ $65$
Find the probability of getting $0, 1, 2$ or $3$ heads.
Verify that the sum of all these probabilities is $1$.
| Number of heads | $0$ | $1$ | $2$ | $3$ |
| Frequency | $70$ | $190$ | $175$ | $65$ |
Find the probability of getting $0, 1, 2$ or $3$ heads.
Verify that the sum of all these probabilities is $1$.
(A-D) Since the coins are tossed $500$ times,the total number of trials is $500$.
Let us denote the events of getting $0$ head,$1$ head,$2$ heads,and $3$ heads by $A_1, A_2, A_3$,and $A_4$ respectively.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$.
$P(A_1) = \frac{70}{500} = 0.14$
$P(A_2) = \frac{190}{500} = 0.38$
$P(A_3) = \frac{175}{500} = 0.35$
$P(A_4) = \frac{65}{500} = 0.13$
Now,verify the sum of all these probabilities:
$P(A_1) + P(A_2) + P(A_3) + P(A_4) = 0.14 + 0.38 + 0.35 + 0.13 = 1$.
Thus,the sum of all probabilities is $1$.
Let us denote the events of getting $0$ head,$1$ head,$2$ heads,and $3$ heads by $A_1, A_2, A_3$,and $A_4$ respectively.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$.
$P(A_1) = \frac{70}{500} = 0.14$
$P(A_2) = \frac{190}{500} = 0.38$
$P(A_3) = \frac{175}{500} = 0.35$
$P(A_4) = \frac{65}{500} = 0.13$
Now,verify the sum of all these probabilities:
$P(A_1) + P(A_2) + P(A_3) + P(A_4) = 0.14 + 0.38 + 0.35 + 0.13 = 1$.
Thus,the sum of all probabilities is $1$.
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