Test by Bikram | Mock GATE | Test 3 | Question: 59

Question

Out of six coins, four coins are tossed simultaneously.

The number of possible outcomes where at most three of the coins turn up as heads is ______.

Comments

The formation of sentence is incorrect and the spelling mistakes there= their ..

---

In the question, in place of almost, atmost will come!!!

---

Question is fixed now. Answer should be ${}^6C_4 *({}^4C_0 + {}^4C_1 + {}^4C_2 + {}^4C_3) = 15 * (1+4+6+4) = 225.$

Answer 1

The question requires you to find number of the outcomes in which at most 3 coins turn up as heads.

i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.

The number of outcomes in which 0 coins turn heads is 6C0 = 1 outcome
The number of outcomes in which 1 coin turns head is  6C1 =  6 outcomes
The number of outcomes in which 2 coins turn heads is 6C2 = 15 outcomes
The number of outcomes in which 3 coins turn heads is 6C0 =  20 outcomes.

Therefore, total number of outcomes =1+6+15+20= 42 outcomes.

Comments

@Arjun sir

In this question won't we consider only 4 coins for the sample space (instead of 6, as only 4 are tossed simultaneously)

---

@Bikram

any update on the this? why won't we are using 6 instead of 4 coins?

---

@logan1x , I still dont understand the question and the answer. If you have got please tell !

---

Both question and answer had problems. Question is fixed now.